Question

Percent by mass of solute (molar mass = 25 g/mol) in its aqueous solution is 30. The mole fraction and molality of the solute in solution are

• A. $Molefraction=0.18;Molality=17.9$
• B. $Molefraction=0.236;Molality=17.14$
• C. $Molefraction=0.236;Molality=15.96$
• D. $Molefraction=0.38;Molality=17.14$

Answer 1

Answer: (B)

Let total mass of solution $=100gm$
Mass of solute $=30gm$ , Mass of solvent (water) $=70gm$
$\therefore molality=\frac{molesofsolute}{massofsolvent(kg)}=\frac{30}{25}\times \frac{10^3}{70}=\frac{1200}{70}=\frac{120}{7}$
$=17.14mol/kg$
Moles of solute $=\frac{30}{25}=\frac{6}{5}mol$
Moles of water $=\frac{70}{18}=\frac{35}{9}mol$
$\therefore molefractionofsolute=\frac{6/5}{\frac{6}{5}+\frac{35}{9}}=\frac{6/5}{\frac{54+175}{45}}=\frac{6/5}{\frac{229}{45}}$
$=\frac{6}{5}\times \frac{45}{229}=\frac{54}{229}=0.236$