Question

The freezing point of a solution containing 50 cm3 of ethylene glycol in 50 g of water is found to be $-{34}^{\circ }C$. Assuming ideal behavior, calculate the density of ethylene glycol $(K_{f}forwater=1.86Kkgmo{l}^{-}1)$.

• A. $2.13gc{m}^{-3}$
• B. $1.13gc{m}^{-3}$
• C. $4.13gc{m}^{-3}$
• D. $3.13gc{m}^{-3}$

Answer 1

The correct option is B $1.13gc{m}^{-3}$

We know where ethylene glycol is used. It acts as anti-freezing agent.
Here we have to calculate density and we know the volume of it.
So how to proceed? Simple.
First we get mass of ethylene glycol from the formula we discussed for depression of freezing point and then calculate density by knowing volume and mass.
So,
$\Delta T=K_f\times \frac{W_2\times 1000}{M{w}_2\times w_1}$
$\Delta T$ = Depression in freezing point
$K_f$ = Molal depression constant of benzene
$W_2$ = Mass of solute
$M{w}_2$ = Molecular mass of solute
$W_1$= Mass of solvent
$34=1.86\times \frac{W_2\times 1000}{62\times 50}$
$W_2=56.66g$
$V=\frac{W_2}{d}$
$50=\frac{56.66}{d}$
$d=1.13gc{m}^{-3}$