Question

The vapour if an aqueous solution of glucose is $750mm$ of Hg at 373 K .Calculate molality of the solution?

• A. $0.523mol/kg$
• B. $0.617mol/kg$
• C. $0.731mol/kg$
• D. $0.836mol/kg$

Answer 1

The correct option is A $0.523mol/kg$

$P_A^o=$ ipper pressure of pure water $760mmHg$

Let mole fraction of Given be $X_B$

$P_A=P_A^o(1-X_B)$

$750=760(1-X_B)$

$1-X_B=\frac{750}{760}$

$X_B=\frac{1}{76}$

Let a moment of water $=1kg$

number of mole $n_A=\frac{1000}{18}$

$X_B=\frac{n_B}{n_A+n_B}{n}_A+n_B\cong n_A$

$\frac{n_B}{n_A}=\frac{1}{76}$

$n_B=\frac{1}{76}\times \frac{1000}{18}=0.731$

molarity $=\frac{n_B}{1kg}=0.731molkg$