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<item xml:lang="es">
		<title>Relationship between molarity and normality (8380)</title>
		<link>https://www.ejercicios-fyq.com/Relationship-between-molarity-and-normality-8380</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Relationship-between-molarity-and-normality-8380</guid>
		<dc:date>2025-01-27T04:52:56Z</dc:date>
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		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Molarity</dc:subject>
		<dc:subject>SOLVED</dc:subject>
		<dc:subject>Normality</dc:subject>

		<description>
&lt;p&gt;Calculate the normality of the following solutions: a) (2 M) b) (0.4 M) c) (3 M) d) (1 M) e) (2 M)&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Calculate the normality of the following solutions: a) &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L54xH20/2da7a629b8ae0d3a94f18434955efb1c-35b24.png?1733007720' style='vertical-align:middle;' width='54' height='20' alt=&#034;\ce{NOH2}&#034; title=&#034;\ce{NOH2}&#034; /&gt; (2 M) b) &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L36xH13/f6fa3f3b19665b50c5e024ec7c43d21c-4bb6a.png?1733037371' style='vertical-align:middle;' width='36' height='13' alt=&#034;\ce{KOH}&#034; title=&#034;\ce{KOH}&#034; /&gt; (0.4 M) c) &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L59xH21/99b33dcab22a7317297ec36909d9f4b5-1645e.png?1734460664' style='vertical-align:middle;' width='59' height='21' alt=&#034;\ce{H2SO_3}&#034; title=&#034;\ce{H2SO_3}&#034; /&gt; (3 M) d) &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L60xH18/cb337c813d41bf56e06ee934b6dd5172-2fb87.png?1733018315' style='vertical-align:middle;' width='60' height='18' alt=&#034;\ce{Al(OH)3}&#034; title=&#034;\ce{Al(OH)3}&#034; /&gt; (1 M) e) &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L44xH15/c1d8eeacb2c44237bbc847837c11be56-d66f1.png?1732958435' style='vertical-align:middle;' width='44' height='15' alt=&#034;\ce{NaCl}&#034; title=&#034;\ce{NaCl}&#034; /&gt; (2 M)&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Normality is defined as the number of equivalents of solute in one liter of solution. The mass of an equivalent is the molecular mass divided by the number of &#034;H&#034; or &#034;OH&#034; groups in the acid or base considered. This means that the relationship between normality and molarity is that normality is equal to molarity multiplied by the number of &#034;H&#034; or &#034;OH&#034; groups in the species. &lt;br/&gt; &lt;br/&gt; a) &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/0738344972da2610cbfb98b6572fd9ed.png' style=&#034;vertical-align:middle;&#034; width=&#034;219&#034; height=&#034;28&#034; alt=&#034;\ce{NOH2}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 4\ N}}&#034; title=&#034;\ce{NOH2}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 4\ N}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; b) &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/231c6ba56c3a66f0d5cb08147fac806e.png' style=&#034;vertical-align:middle;&#034; width=&#034;246&#034; height=&#034;28&#034; alt=&#034;\ce{KOH}\ (0.4\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 0.4\ N}}&#034; title=&#034;\ce{KOH}\ (0.4\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 0.4\ N}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; c) &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/316e09d835185628eb3f2872f7045160.png' style=&#034;vertical-align:middle;&#034; width=&#034;224&#034; height=&#034;28&#034; alt=&#034;\ce{H_2SO_3}\ (3\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 6\ N}}&#034; title=&#034;\ce{H_2SO_3}\ (3\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 6\ N}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; d) &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/b2dbaf95dd7ea26cea3bf959928e2faa.png' style=&#034;vertical-align:middle;&#034; width=&#034;243&#034; height=&#034;28&#034; alt=&#034;\ce{Al(OH)_3}\ (1\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 3\ N}}&#034; title=&#034;\ce{Al(OH)_3}\ (1\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 3\ N}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; e) &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/c83eb18dd999260e0f60514be891e947.png' style=&#034;vertical-align:middle;&#034; width=&#034;210&#034; height=&#034;28&#034; alt=&#034;\ce{NaCl}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 2\ N}}&#034; title=&#034;\ce{NaCl}\ (2\ M)\ \to\ \fbox{\color[RGB]{192,0,0}{\bf 2\ N}}&#034; /&gt; (because it is a salt with a 1:1 stoichiometry between its ions).&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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	</item>
<item xml:lang="es">
		<title>Molarity from % (m/V) concentration (8303)</title>
		<link>https://www.ejercicios-fyq.com/Molarity-from-m-V-concentration-8303</link>
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		<dc:date>2024-09-08T03:42:27Z</dc:date>
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		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Concentration</dc:subject>
		<dc:subject>Molarity</dc:subject>
		<dc:subject>Percentage (mass/volume)</dc:subject>
		<dc:subject>Solutions</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;Calculate the molarity of a sulfurous acid solution whose concentration is (m/V).&lt;/p&gt;


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&lt;a href="https://www.ejercicios-fyq.com/Percentage-mass-volume" rel="tag"&gt;Percentage (mass/volume)&lt;/a&gt;, 
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		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Calculate the molarity of a sulfurous acid solution whose concentration is &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L30xH19/86ee91cf864173a2378fbdeb1f3d916e-a72b9.png?1732971539' style='vertical-align:middle;' width='30' height='19' alt=&#034;8\ \%&#034; title=&#034;8\ \%&#034; /&gt; (m/V).&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;First, set a quantity of solution as the basis for calculation. Consider 100 mL of solution, which gives you 8 g of solute, &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/ebddc00967ec960653ac0c88f6e17e8b.png' style=&#034;vertical-align:middle;&#034; width=&#034;46&#034; height=&#034;16&#034; alt=&#034;\ce{H2SO3}&#034; title=&#034;\ce{H2SO3}&#034; /&gt;, (these amounts are the &#8220;translation&#8221; of &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/86ee91cf864173a2378fbdeb1f3d916e.png' style=&#034;vertical-align:middle;&#034; width=&#034;30&#034; height=&#034;19&#034; alt=&#034;8\ \%&#034; title=&#034;8\ \%&#034; /&gt; m/V). &lt;br/&gt; &lt;br/&gt; The moles corresponding to that mass of solute are obtained from the molecular mass of the acid: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/cebd2432beffb5b61436a43762fc2357.png' style=&#034;vertical-align:middle;&#034; width=&#034;384&#034; height=&#034;44&#034; alt=&#034;M_{\ce{H2SO3}} = 2\cdot 1 + 1\cdot 32 + 3\cdot 16 = \color[RGB]{0,112,192}{\bm{82\ \frac{g}{mol}}}&#034; title=&#034;M_{\ce{H2SO3}} = 2\cdot 1 + 1\cdot 32 + 3\cdot 16 = \color[RGB]{0,112,192}{\bm{82\ \frac{g}{mol}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Calculate the moles equivalent to the mass of solute: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/2474f672e60f0538a05a977d56c8f8dc.png' style=&#034;vertical-align:middle;&#034; width=&#034;353&#034; height=&#034;51&#034; alt=&#034;8\ \cancel{g}\ \ce{H2SO3}\cdot \frac{1\ mol}{82\ \cancel{g}} = \color[RGB]{0,112,192}{\bm{9.76\cdot 10^{-2}\ mol}}&#034; title=&#034;8\ \cancel{g}\ \ce{H2SO3}\cdot \frac{1\ mol}{82\ \cancel{g}} = \color[RGB]{0,112,192}{\bm{9.76\cdot 10^{-2}\ mol}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The molarity is the ratio between the moles of solute and the volume of the solution, expressed in liters, i.e., 0.1 L (because you considered 100 mL at the beginning): &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/30c4ea39edaee0d5ecc8e65e6338aed1.png' style=&#034;vertical-align:middle;&#034; width=&#034;318&#034; height=&#034;48&#034; alt=&#034;M = \frac{9.76\cdot 10^{-2}\ mol}{0.1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{0.98\ \frac{mol}{L}}}}&#034; title=&#034;M = \frac{9.76\cdot 10^{-2}\ mol}{0.1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{0.98\ \frac{mol}{L}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Molar fraction and molality of a solution (8298)</title>
		<link>https://www.ejercicios-fyq.com/Molar-fraction-and-molality-of-a-solution-8298</link>
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		<dc:date>2024-09-03T07:07:35Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Concentration</dc:subject>
		<dc:subject>Molar fraction</dc:subject>
		<dc:subject>Molality</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;A solution contains by mass of HCl: &lt;br class='autobr' /&gt;
a) Calculate the molar fraction of HCl. &lt;br class='autobr' /&gt;
b) Calculate the molality of HCl in the solution.&lt;/p&gt;


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&lt;a href="https://www.ejercicios-fyq.com/Molality" rel="tag"&gt;Molality&lt;/a&gt;, 
&lt;a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag"&gt;SOLVED&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;A solution contains &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L41xH19/2a4c30361eca926f0aaeb2da13868197-4e2d5.png?1733054802' style='vertical-align:middle;' width='41' height='19' alt=&#034;36\ \%&#034; title=&#034;36\ \%&#034; /&gt; by mass of HCl:&lt;/p&gt;
&lt;p&gt;a) Calculate the molar fraction of HCl.&lt;/p&gt;
&lt;p&gt;b) Calculate the molality of HCl in the solution.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;You can fix an amount of 100 g of solution to solve the problem because, in those 100 g of solution, there will be 36 g of HCl (which is what the percentage means) and 64 g of water. Convert both quantities to moles: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/a49a404ba95e3063c6c49ca14f62bc30.png' style=&#034;vertical-align:middle;&#034; width=&#034;340&#034; height=&#034;51&#034; alt=&#034;36\ \cancel{g}\ \ce{HCl}\cdot \frac{1\ mol}{36.5\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{0.986\ \ce{mol\ HCl}}}&#034; title=&#034;36\ \cancel{g}\ \ce{HCl}\cdot \frac{1\ mol}{36.5\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{0.986\ \ce{mol\ HCl}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/c3f6a1771f1324c8967fd65cf19a949a.png' style=&#034;vertical-align:middle;&#034; width=&#034;344&#034; height=&#034;51&#034; alt=&#034;64\ \cancel{g}\ \ce{H2O}\cdot \frac{1\ mol}{18\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{3.556\ \ce{mol\ H2O}}}&#034; title=&#034;64\ \cancel{g}\ \ce{H2O}\cdot \frac{1\ mol}{18\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{3.556\ \ce{mol\ H2O}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; a) Calculate the molar fraction: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/57a80ae804207b5ac9dc81a36c1a73f4.png' style=&#034;vertical-align:middle;&#034; width=&#034;493&#034; height=&#034;53&#034; alt=&#034;x_{\ce{HCl}}= \frac{n_{\ce{HCl}}}{n_{\ce{HCl}} + n_{\ce{H2O}}} = \frac{0.986\ \cancel{mol}}{(0.986 + 3.556)\ \cancel{mol}} = \fbox{\color[RGB]{192,0,0}{\bf 0.217}}&#034; title=&#034;x_{\ce{HCl}}= \frac{n_{\ce{HCl}}}{n_{\ce{HCl}} + n_{\ce{H2O}}} = \frac{0.986\ \cancel{mol}}{(0.986 + 3.556)\ \cancel{mol}} = \fbox{\color[RGB]{192,0,0}{\bf 0.217}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Calculate the molality. This is defined as the ratio between the moles of solute and the mass of solvent, expressed in kilograms: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/697a9f7f7aecfa89964b16925166bec2.png' style=&#034;vertical-align:middle;&#034; width=&#034;421&#034; height=&#034;50&#034; alt=&#034;m = \frac{n_{\ce{HCl}}}{m_{\ce{H2O}}\ (kg)} = \frac{0.986\ mol}{6.4\cdot 10^{-2}\ kg}= \fbox{\color[RGB]{192,0,0}{\bm{15.4\ \frac{mol}{kg}}}}&#034; title=&#034;m = \frac{n_{\ce{HCl}}}{m_{\ce{H2O}}\ (kg)} = \frac{0.986\ mol}{6.4\cdot 10^{-2}\ kg}= \fbox{\color[RGB]{192,0,0}{\bm{15.4\ \frac{mol}{kg}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Raoult's law and vapor pressure (8282)</title>
		<link>https://www.ejercicios-fyq.com/Raoult-s-law-and-vapor-pressure-8282</link>
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		<dc:date>2024-08-09T03:43:44Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Mole fraction</dc:subject>
		<dc:subject>Partial pressure</dc:subject>
		<dc:subject>Colligative properties</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;Calculate the vapor pressure at of a solution containing 150 grams of glucose dissolved in 140 grams of ethyl alcohol. The vapor pressure of ethyl alcohol at is 43 mm Hg.&lt;/p&gt;


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&lt;a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag"&gt;SOLVED&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Calculate the vapor pressure at &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L55xH42/600047c1d05cb8002892c032010910ad-51014.png?1732969135' style='vertical-align:middle;' width='55' height='42' alt=&#034;20\ ^oC&#034; title=&#034;20\ ^oC&#034; /&gt; of a solution containing 150 grams of glucose dissolved in 140 grams of ethyl alcohol. The vapor pressure of ethyl alcohol at &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L55xH42/600047c1d05cb8002892c032010910ad-51014.png?1732969135' style='vertical-align:middle;' width='55' height='42' alt=&#034;20\ ^oC&#034; title=&#034;20\ ^oC&#034; /&gt; is 43 mm Hg.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Glucose is &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/c283034f9003b340ebe061f23f1b1584.png' style=&#034;vertical-align:middle;&#034; width=&#034;61&#034; height=&#034;16&#034; alt=&#034;\ce{C6H12O6}&#034; title=&#034;\ce{C6H12O6}&#034; /&gt; and has a molecular mass of 180 g/mol. Ethyl alcohol is &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/2765b945cf026723953ab968517ccd49.png' style=&#034;vertical-align:middle;&#034; width=&#034;85&#034; height=&#034;16&#034; alt=&#034;\ce{CH3CH2OH}&#034; title=&#034;\ce{CH3CH2OH}&#034; /&gt; and has a molecular mass of 46 g/mol. Using this data, calculate the moles of each substance and determine the mole fraction of alcohol in the mixture. &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/fd9d7256ed701275e28a18d4b2a24909.png' style=&#034;vertical-align:middle;&#034; width=&#034;437&#034; height=&#034;52&#034; alt=&#034;150\ \cancel{g}\ \ce{C6H12O6}\cdot \frac{1\ \text{mol}}{180\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{0.833\ mol\ \ce{C6H12O6}}}&#034; title=&#034;150\ \cancel{g}\ \ce{C6H12O6}\cdot \frac{1\ \text{mol}}{180\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{0.833\ mol\ \ce{C6H12O6}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/8459314dac433cb5b7765e119baad222.png' style=&#034;vertical-align:middle;&#034; width=&#034;403&#034; height=&#034;52&#034; alt=&#034;140\ \cancel{g}\ \ce{C2H6O}\cdot \frac{1\ \text{mol}}{46\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{3.043\ mol\ \ce{C2H6O}}}&#034; title=&#034;140\ \cancel{g}\ \ce{C2H6O}\cdot \frac{1\ \text{mol}}{46\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{3.043\ mol\ \ce{C2H6O}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The mole fraction of alcohol is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/d684317ff9259ce1be15344d83f7b7b2.png' style=&#034;vertical-align:middle;&#034; width=&#034;555&#034; height=&#034;53&#034; alt=&#034;x_{\ce{C2H6O}} = \frac{n_{\ce{C2H6O}}}{n_{\ce{C2H6O}} + n_{\ce{C6H12O6}}} = \frac{3.043\ \cancel{\text{mol}}}{(0.833 + 3.043)\ \cancel{\text{mol}}} = \color[RGB]{0,112,192}{\bf 0.785}&#034; title=&#034;x_{\ce{C2H6O}} = \frac{n_{\ce{C2H6O}}}{n_{\ce{C2H6O}} + n_{\ce{C6H12O6}}} = \frac{3.043\ \cancel{\text{mol}}}{(0.833 + 3.043)\ \cancel{\text{mol}}} = \color[RGB]{0,112,192}{\bf 0.785}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Now calculate the vapor pressure using Raoult's law: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/cd3edf3caa7b39581176da0db7a77539.png' style=&#034;vertical-align:middle;&#034; width=&#034;568&#034; height=&#034;32&#034; alt=&#034;P_{\ce{C2H6O}} = x_{\ce{C2H6O}}\cdot P_T = 0.785\cdot 43\ \text{mm Hg} = \fbox{\color[RGB]{192,0,0}{\textbf{33.75\ \ce{mm\ Hg}}}}&#034; title=&#034;P_{\ce{C2H6O}} = x_{\ce{C2H6O}}\cdot P_T = 0.785\cdot 43\ \text{mm Hg} = \fbox{\color[RGB]{192,0,0}{\textbf{33.75\ \ce{mm\ Hg}}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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	</item>
<item xml:lang="es">
		<title>Mass of a volume of ethanol, expressed by pounds (4883)</title>
		<link>https://www.ejercicios-fyq.com/Mass-of-a-volume-of-ethanol-expressed-by-pounds-4883</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Mass-of-a-volume-of-ethanol-expressed-by-pounds-4883</guid>
		<dc:date>2019-01-03T08:28:18Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Concentration</dc:subject>
		<dc:subject>Units conversion</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;The density of ethanol is 0.79 g/mL. Calculate the mass, expressed in pounds, of 17.4 mL of ethanol, knowing that one pound is equal to 0.45 kg. Express the result using scientific notation.&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;The density of ethanol is 0.79 g/mL. Calculate the mass, expressed in pounds, of 17.4 mL of ethanol, knowing that one pound is equal to 0.45 kg. Express the result using scientific notation.&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;We will solve this exercise in two steps. First, we will use the definition of density to calculate the mass of ethanol. Second, we will convert the units. &lt;br/&gt; &lt;br/&gt; 1. Calculate the mass of ethanol: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/6d800afacacbd8edf9ada35bbb85865b.png' style=&#034;vertical-align:middle;&#034; width=&#034;584&#034; height=&#034;42&#034; alt=&#034;d = \frac{m}{V}\ \to\ {\color[RGB]{2,112,20}{\bm{m = d\cdot V}}}\ \to\ m = 0.79\ \frac{g}{\cancel{mL}}\cdot 17.4\ \cancel{mL} = \color[RGB]{0,112,192}{\bf 13.75\ g}&#034; title=&#034;d = \frac{m}{V}\ \to\ {\color[RGB]{2,112,20}{\bm{m = d\cdot V}}}\ \to\ m = 0.79\ \frac{g}{\cancel{mL}}\cdot 17.4\ \cancel{mL} = \color[RGB]{0,112,192}{\bf 13.75\ g}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; 2. Convert the mass to pounds: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/79ead6abd97a6c6ae1780f372cf14112.png' style=&#034;vertical-align:middle;&#034; width=&#034;393&#034; height=&#034;59&#034; alt=&#034;13.75\ \cancel{g}\cdot \frac{1\ \cancel{kg}}{10^3\ \cancel{g}}\cdot \frac{1\ lb}{0.45\ \cancel{kg}} = \fbox{\color[RGB]{192,0,0}{\bm{3.06\cdot 10^{-2}\ lb}}}&#034; title=&#034;13.75\ \cancel{g}\cdot \frac{1\ \cancel{kg}}{10^3\ \cancel{g}}\cdot \frac{1\ lb}{0.45\ \cancel{kg}} = \fbox{\color[RGB]{192,0,0}{\bm{3.06\cdot 10^{-2}\ lb}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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	</item>
<item xml:lang="es">
		<title>Solutions: Percentage by mass (1931)</title>
		<link>https://www.ejercicios-fyq.com/Solutions-Percentage-by-mass-1931</link>
		<guid isPermaLink="true">https://www.ejercicios-fyq.com/Solutions-Percentage-by-mass-1931</guid>
		<dc:date>2012-11-10T07:55:39Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Solutions</dc:subject>
		<dc:subject>Percentage by mass</dc:subject>
		<dc:subject>SOLVED</dc:subject>

		<description>
&lt;p&gt;Make a solution by mixing 22 g of sugar and 240 mL of milk. Determine the percentage by mass of this solution. How much sugar is needed to prepare 500 g of solution with the same concentration? &lt;br class='autobr' /&gt;
Average density of milk is .&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Make a solution by mixing 22 g of sugar and 240 mL of milk. Determine the percentage by mass of this solution. How much sugar is needed to prepare 500 g of solution with the same concentration?&lt;/p&gt;
&lt;p&gt;Average density of milk is &lt;img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L121xH24/2206c99476a92d0e3dd8df6c47e6f879-3757a.png?1732956019' style='vertical-align:middle;' width='121' height='24' alt=&#034;1.03\ g\cdot mL^{-1}&#034; title=&#034;1.03\ g\cdot mL^{-1}&#034; /&gt;.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;The mass of milk used for the mixture is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/1c73770d1fdf774e1bbc902ec655d4ad.png' style=&#034;vertical-align:middle;&#034; width=&#034;248&#034; height=&#034;47&#034; alt=&#034;240\ \cancel{mL}\cdot \frac{1.03\ g}{1\ \cancel{mL}} = \color[RGB]{0,112,192}{\bf 247.2\ g}&#034; title=&#034;240\ \cancel{mL}\cdot \frac{1.03\ g}{1\ \cancel{mL}} = \color[RGB]{0,112,192}{\bf 247.2\ g}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The total mass of the solution is: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/fd21c5fc2aaa88031f7651d4930c3e9f.png' style=&#034;vertical-align:middle;&#034; width=&#034;482&#034; height=&#034;23&#034; alt=&#034;m_T = m_S + m_d = (22 + 247.2)\ g\ \to\ m_D = \color[RGB]{0,112,192}{\bf 269.2\ g}&#034; title=&#034;m_T = m_S + m_d = (22 + 247.2)\ g\ \to\ m_D = \color[RGB]{0,112,192}{\bf 269.2\ g}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; The mass percentage is: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/38cb9992a9391a5d6a27803de6e81f53.png' style=&#034;vertical-align:middle;&#034; width=&#034;416&#034; height=&#034;51&#034; alt=&#034;\%(m) = \frac{m_S}{m_T}\cdot 100 = \frac{22\ \cancel{g}}{269.2\ \cancel{g}}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bf 8.17\ \%}}&#034; title=&#034;\%(m) = \frac{m_S}{m_T}\cdot 100 = \frac{22\ \cancel{g}}{269.2\ \cancel{g}}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bf 8.17\ \%}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; The obtained result indicates that you need 8.17 g of sugar to prepare 100 g of solution: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://www.ejercicios-fyq.com/local/cache-TeX/e25d3d805fa0d6cf5030bbc04c5e644e.png' style=&#034;vertical-align:middle;&#034; width=&#034;297&#034; height=&#034;52&#034; alt=&#034;500\ \cancel{g\ D}\cdot \frac{8.17\ g\ S}{100\ \cancel{g\ D}} = \fbox{\color[RGB]{192,0,0}{\bf 40.9\ g\ S}}&#034; title=&#034;500\ \cancel{g\ D}\cdot \frac{8.17\ g\ S}{100\ \cancel{g\ D}} = \fbox{\color[RGB]{192,0,0}{\bf 40.9\ g\ S}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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