Question

A solution containing $ 2.44\,g $ of a solute dissolved in $ 75\,g $ of water boiled at $ 100.413^{\circ}C $ . What will be the molar mass of the solute
$ (K_b $ for water $ = 0.52 \,K \,kg\, mol^{-1}) $ ?

• A. $ 40.96\, g\, mol^{-1} $
• B. $ 20.48\, g\, mol^{-1} $
• C. $ 81.92\, g\, mol^{-1} $
• D. None of the above

Answer 1

Answer: (A)

$W_{\text{solute}} = 2.44\,g$
$W_{\text{solvent}} = 75 \,g$
$T_b = 100.413^{\circ} \,C$
$\therefore \Delta T_b = 100.413 - 100^{\circ} C$
$= 0.413^{\circ}C$
Thus, $M_A = \frac{K_b \times W_{\text{solute}} \times 1000}{\Delta T_b \times W_{\text{solvent}}}$
$ = \frac{0.52 \times 2.44 \times 1000}{75\times 0.413}$
$ = 40 . 96 \,g \,mol^{-1}$