Question

A solution containing $3.24$ of a nonvolatile nonelectrolyte and $200g$ of water boils at ${100.130}^{\circ }C$ at 1 atm. The molecular weight of the solute is:
($K_b$ for water ${0.513}^{\circ }$C/m)

• A. $64$ g/mol
• B. $32$ g/mol
• C. $16$ g/mol
• D. $128$ g/mol

Answer 1

The correct option is A $64$ g/mol

Let $M$ be the molecular weight of solute.

The number of moles is the ratio of mass to molecular weight.

Number of moles of solute $=\frac{3.24}{M}$

The molality of solution is the number of moles of solute in $1000$ g of water.

Molality $=m=\frac{3.24\times 1000}{M\times 200}=\frac{16.2}{M}$

Elevation in the boiling point $=\Delta T_b=100.130-100={0.130}^{o}C$

$\Delta T_b=K_{b}m$

$0.130=0.513\times \frac{16.2}{M}$

$M=\frac{0.513\times 16.2}{0.130}=64g/mol$