Question

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

• A. $40.96gmo{l}^{-1}$
• B. $20.48gmo{l}^{-1}$
• C. $81.92gmo{l}^{-1}$
• D. None of the above

Answer 1

Answer: (A)

$W_{\text{solute}}=2.44g$
$W_{\text{solvent}}=75g$
$T_b=100.413^{\circ }C$
$\therefore \Delta T_b=100.413-100^{\circ }C$
$=0.413^{\circ }C$
Thus, $M_A=\frac{K_b\times W_{\text{solute}}\times 1000}{\Delta T_b\times W_{\text{solvent}}}$
$=\frac{0.52\times 2.44\times 1000}{75\times 0.413}$
$=40.96gmo{l}^{-1}$