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://ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://ejercicios-fyq.com/Molar-fraction-and-molality-of-a-solution-8298 https://ejercicios-fyq.com/Molar-fraction-and-molality-of-a-solution-8298 2024-09-03T07:07:35Z text/html es F_y_Q Concentration Molar fraction Molality SOLVED <p>A solution contains by mass of HCl: <br class='autobr' /> a) Calculate the molar fraction of HCl. <br class='autobr' /> b) Calculate the molality of HCl in the solution.</p> - <a href="https://ejercicios-fyq.com/Solutions-271" rel="directory">Solutions</a> / <a href="https://ejercicios-fyq.com/Concentration" rel="tag">Concentration</a>, <a href="https://ejercicios-fyq.com/Molar-fraction" rel="tag">Molar fraction</a>, <a href="https://ejercicios-fyq.com/Molality" rel="tag">Molality</a>, <a href="https://ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>A solution contains <img src='https://ejercicios-fyq.com/local/cache-vignettes/L41xH19/2a4c30361eca926f0aaeb2da13868197-4e2d5.png?1733054802' style='vertical-align:middle;' width='41' height='19' alt="36\ \%" title="36\ \%" /> by mass of HCl:</p> <p>a) Calculate the molar fraction of HCl.</p> <p>b) Calculate the molality of HCl in the solution.</math></p></div> <hr /> <div <div class='rss_ps'><p>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: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/a49a404ba95e3063c6c49ca14f62bc30.png' style="vertical-align:middle;" width="340" height="51" alt="36\ \cancel{g}\ \ce{HCl}\cdot \frac{1\ mol}{36.5\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{0.986\ \ce{mol\ HCl}}}" title="36\ \cancel{g}\ \ce{HCl}\cdot \frac{1\ mol}{36.5\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{0.986\ \ce{mol\ HCl}}}" /> <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/c3f6a1771f1324c8967fd65cf19a949a.png' style="vertical-align:middle;" width="344" height="51" alt="64\ \cancel{g}\ \ce{H2O}\cdot \frac{1\ mol}{18\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{3.556\ \ce{mol\ H2O}}}" title="64\ \cancel{g}\ \ce{H2O}\cdot \frac{1\ mol}{18\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{3.556\ \ce{mol\ H2O}}}" /> <br/> <br/> a) Calculate the molar fraction: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/57a80ae804207b5ac9dc81a36c1a73f4.png' style="vertical-align:middle;" width="493" height="53" alt="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}}" title="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}}" /></p> <br/> b) Calculate the molality. This is defined as the ratio between the moles of solute and the mass of solvent, expressed in kilograms: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/697a9f7f7aecfa89964b16925166bec2.png' style="vertical-align:middle;" width="421" height="50" alt="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}}}}" title="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}}}}" /></p> </math></p></div>