https://ejercicios-fyq.com/ Ejercicios Resueltos, Situaciones de aprendizaje y VÍDEOS de Física y Química para Secundaria y Bachillerato es SPIP - www.spip.net https://www.ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://www.ejercicios-fyq.com/Raoult-s-law-and-vapor-pressure-8282 https://www.ejercicios-fyq.com/Raoult-s-law-and-vapor-pressure-8282 2024-08-09T03:43:44Z text/html es F_y_Q Mole fraction Partial pressure Colligative properties SOLVED <p>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.</p> - <a href="https://www.ejercicios-fyq.com/Solutions-271" rel="directory">Solutions</a> / <a href="https://www.ejercicios-fyq.com/Mole-fraction" rel="tag">Mole fraction</a>, <a href="https://www.ejercicios-fyq.com/Partial-pressure" rel="tag">Partial pressure</a>, <a href="https://www.ejercicios-fyq.com/Colligative-properties" rel="tag">Colligative properties</a>, <a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>Calculate the vapor pressure at <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L55xH42/600047c1d05cb8002892c032010910ad-51014.png?1732969135' style='vertical-align:middle;' width='55' height='42' alt="20\ ^oC" title="20\ ^oC" /> of a solution containing 150 grams of glucose dissolved in 140 grams of ethyl alcohol. The vapor pressure of ethyl alcohol at <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L55xH42/600047c1d05cb8002892c032010910ad-51014.png?1732969135' style='vertical-align:middle;' width='55' height='42' alt="20\ ^oC" title="20\ ^oC" /> is 43 mm Hg.</math></p></div> <hr /> <div <div class='rss_ps'><p>Glucose is <img src='https://www.ejercicios-fyq.com/local/cache-TeX/c283034f9003b340ebe061f23f1b1584.png' style="vertical-align:middle;" width="61" height="16" alt="\ce{C6H12O6}" title="\ce{C6H12O6}" /> and has a molecular mass of 180 g/mol. Ethyl alcohol is <img src='https://www.ejercicios-fyq.com/local/cache-TeX/2765b945cf026723953ab968517ccd49.png' style="vertical-align:middle;" width="85" height="16" alt="\ce{CH3CH2OH}" title="\ce{CH3CH2OH}" /> 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. <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/fd9d7256ed701275e28a18d4b2a24909.png' style="vertical-align:middle;" width="437" height="52" alt="150\ \cancel{g}\ \ce{C6H12O6}\cdot \frac{1\ \text{mol}}{180\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{0.833\ mol\ \ce{C6H12O6}}}" title="150\ \cancel{g}\ \ce{C6H12O6}\cdot \frac{1\ \text{mol}}{180\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{0.833\ mol\ \ce{C6H12O6}}}" /> <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/8459314dac433cb5b7765e119baad222.png' style="vertical-align:middle;" width="403" height="52" alt="140\ \cancel{g}\ \ce{C2H6O}\cdot \frac{1\ \text{mol}}{46\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{3.043\ mol\ \ce{C2H6O}}}" title="140\ \cancel{g}\ \ce{C2H6O}\cdot \frac{1\ \text{mol}}{46\ \cancel{g}} = \color[RGB]{0,112,192}{\textbf{3.043\ mol\ \ce{C2H6O}}}" /> <br/> <br/> The mole fraction of alcohol is: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/d684317ff9259ce1be15344d83f7b7b2.png' style="vertical-align:middle;" width="555" height="53" alt="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}" title="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}" /> <br/> <br/> Now calculate the vapor pressure using Raoult's law: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/cd3edf3caa7b39581176da0db7a77539.png' style="vertical-align:middle;" width="568" height="32" alt="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}}}}" title="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}}}}" /></p> </math></p></div>