Question

The vapour pressure of water is $92$mm at $323$K. $18.1$g of urea are dissolved in $100$g of water. The vapour pressure is reduced by $5$mm. Calculate the molar mass of urea.

Answer 1

Given: $P\stackrel{o}{{}_A}=92mm,P\stackrel{o}{{}_A}-P_A=5mm$

$W_A=100,W_B=18.1,$

$M_A=18$ $M_B=?$ $n_A=\frac{W_A}{M_A}=\frac{100}{18}=5.56$

$n_B=\frac{W_B}{M_B}=\frac{18.1}{M_B}$

$\Rightarrow \frac{P\stackrel{o}{{}_A}-P_A}{P\stackrel{o}{{}_A}}=\frac{n_B}{n_A+n_B}$

$\Rightarrow \frac{5}{92}=\frac{18.1/M_B}{5.56+\frac{18.1}{M_B}}$

$\Rightarrow 0.054(5.56+\frac{18.1}{M_B})=\frac{18.1}{M_B}$

$\Rightarrow 0.3+\frac{0.97}{M_B}=\frac{18.1}{M_B}$

$\Rightarrow 0.3=\frac{17.73}{M_B}\Rightarrow M_B=\frac{17.73}{0.3}=57.1$

$\therefore$ Molar mass of Urea= $57.1g/mol$