Question

The freezing point of a solution $50c{m}^3$ of ethylene glycol in $50g$ water is found to be ${-34}^{\circ }C$. Assuming ideal behaviour, calculate the density of ethylene glycol$(K_{f}for{H}_{2}O=1.86Kkgmo{l}^{-1}$)
Given:
$\text{Molar mass of Ethylene glycol}=62gmo{l}^{-1}$

• A. $3.333gc{m}^{-3}$
• B. $0.667gc{m}^{-3}$
• C. $3.554gc{m}^{-3}$
• D. $1.133gc{m}^{-3}$

Answer 1

The correct option is D $1.133gc{m}^{-3}$

$\text{Ethylene glycol}=50c{m}^3$

$K_{f}forwater=1.86Kkgmo{l}^{-1}$

$\text{Mass of ethylene glycol (w)}=V\times \rho =50\times \rho$

$\text{Mass of water (W)}=50g$

$\Delta T_f=0-(-34)=+{34}^{\circ }C$
$\Delta T_f=K_f\times m$
Here $m$ is molality

$\Delta T_f=\frac{K_f\times w\times 1000}{M_{\text{Ethylene glycol}}\times W}$

$34=\frac{1.86\times 50\times \rho \times 1000}{62\times 50}$

$\rho =1.133gc{m}^{-3}$