Question

An aqueous solution containing $288g$ of a non-volatile compound having the stoichimetric composition $C_x{H}_{2x}{O}_x$ in $90g$ water boils at ${101.24}^{\circ }C$ at one atmospheric pressure. What is the molecular formula of the compound?
$K_b\left(H_{2}O\right)=0.512Kmo{l}^{-1}kg,T_b\left(H_{2}O\right)={100}^{\circ }C$

• A. $C_{44}{H}_{88}{O}_{44}$
• B. $C_{12}{H}_{24}{O}_{12}$
• C. $C_{22}{H}_{44}{O}_{22}$
• D. $C_{10}{H}_{20}{O}_{10}$

Answer 1

The correct option is A $C_{44}{H}_{88}{O}_{44}$

Boiling point elevation of a solution is given by,
$△T_b=K_b\times m$
where,
$△T_b$ is the elevation in boiling point.
$K_b$ is molal elevation constant.
m is molality of solution.

Given:
$\Delta T={1.24}^{o}C$
$K_b\left(H_{2}O\right)=0.512Kmo{l}^{-1}kg$
Mass of solute $=288g$
Mass of solvent $=90g$
Molar mass of solute is 'M'

To calculate the molecular mass of
$C_x{H}_{2x}{O}_x$

$△T_f=\frac{K_f\times \frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$

$1.24=\left(0.512\right)\times \frac{\frac{288}{M}}{90}\times 1000$

$M=\frac{\left(288\right)\left(1000\right)\left(0.512\right)}{\left(90\right)\left(1.24\right)}$

$M=1321.2g$

To calculate the molecular formula of
$C_x{H}_{2x}{O}_x$
$12x+2x+16x=1321.1$

$x=44$
$\therefore$
$C_{44}{H}_{88}{O}_{44}$