Question

If a automobile radiator holds 1.0 Kg of water; how many grams of ethylene glycol $\left(C_2{H}_6{O}_2\right)$ must be add to get the freezing point of the solution lowered to $-{2.79}^{\circ }C$?
[$K_f$ for water is $1.86Kkgmo{l}^{-1}$]

• A. 93 gm
• B. 39 gm
• C. 27 gm
• D. 72 gm

Answer 1

The correct option is A 93 gm

$K_f=1.86Kkgmo{l}^{-1}$
$\Delta T_f=0-(-2.79)={2.79}^{\circ }C$
Mass of solvent = 1.0 Kg
Mass of solute = ?
Molecular mass of solute = 62 g/mol
$\Delta T_f=K_f\times m$
$m=\frac{\frac{\text{weight of solute}}{\text{molecular mass of solute}}}{\text{mass of solvent in g}}\times 1000$
$m=\frac{W/62}{1000}\times 1000=W/62$
$2.79=1.86\times \frac{W}{62}\Rightarrow W=\frac{62\times 2.79}{1.86}=93gm$