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Q.
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
NTA AbhyasNTA Abhyas 2020Solutions
Solution:
Let total mass of solution $=100 \, gm$
Mass of solute $=30 \, gm$ , Mass of solvent (water) $=70 \, gm$
$\therefore molality=\frac{moles of solute}{mass of solvent \left(\right. kg \left.\right)}=\frac{30}{25}\times \, \frac{1 0^{3}}{70}=\frac{1200}{70}=\frac{120}{7}$
$=17.14mol/kg$
Moles of solute $=\frac{30}{25}=\frac{6}{5}mol$
Moles of water $=\frac{70}{18}=\frac{35}{9}mol$
$\therefore mole \, fractionofsolute=\frac{6 / 5}{\frac{6}{5} + \frac{35}{9}}=\frac{6 / 5}{\frac{54 + 175}{45}}=\frac{6 / 5}{\frac{229}{45}}$
$=\frac{6}{5} \, \times \, \frac{45}{229}=\frac{54}{229}=0.236$