Question

$83g$ of ethylene glycol dissolved in $625g$ of water.
The freezing point of the solution is $K.$ (Nearest integer)
[Use: Molal Freezing point depression constant of water $=1.86Kkgmo{l}^{-1}$ ]
Freezing Point of water $=273K$
Atomic masses : $C:12.0u,O:16.0u,H:1.0u$ ]

Answer 1

Answer: 269

$k_f=1.86k.kg/mol$
$T_f^{\circ }=273k$
solvent $:H_{2}O(625g)$

solute
$\Rightarrow \Delta T_f=k_f\times m$
$\Rightarrow \left(T_f^{\circ }-T_f^1\right)=1.86\times \frac{83/62}{624/1000}$
$\Rightarrow 273-T_f^1=\frac{1.86\times 83\times 1000}{62\times 625}$
$=\frac{154380}{38750}$
$\Rightarrow 273-T_f^1=4$
$\Rightarrow T_f^1=259K$