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Molar fraction and molality of a solution (8298)
Martes 3 de septiembre de 2024, por
A solution contains 
by mass of HCl:
a) Calculate the molar fraction of HCl.
b) Calculate the molality of HCl in the solution.
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:
![36\ \cancel{g}\ \ce{HCl}\cdot \frac{1\ mol}{36.5\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{0.986\ \ce{mol\ HCl}}}](local/cache-vignettes/L340xH51/a49a404ba95e3063c6c49ca14f62bc30-51215.png?1733072531 "36\ \cancel{g}\ \ce{HCl}\cdot \frac{1\ mol}{36.5\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{0.986\ \ce{mol\ HCl}}}")
![64\ \cancel{g}\ \ce{H2O}\cdot \frac{1\ mol}{18\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{3.556\ \ce{mol\ H2O}}}](local/cache-vignettes/L344xH51/c3f6a1771f1324c8967fd65cf19a949a-c9c2a.png?1733072531 "64\ \cancel{g}\ \ce{H2O}\cdot \frac{1\ mol}{18\ \cancel{g}}= \color[RGB]{0,112,192}{\textbf{3.556\ \ce{mol\ H2O}}}")
a) Calculate the molar fraction:
![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}}](local/cache-vignettes/L493xH53/57a80ae804207b5ac9dc81a36c1a73f4-acf8a.png?1733072531 "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}}")
b) Calculate the molality. This is defined as the ratio between the moles of solute and the mass of solvent, expressed in kilograms:
![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}}}}](local/cache-vignettes/L421xH50/697a9f7f7aecfa89964b16925166bec2-fa316.png?1733072531 "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}}}}")