1. Tardigrade
  2. Question
  3. Chemistry
  4. 83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is K . (Nearest integer) [Use: Molal Freezing point depression constant of water =1.86 K kg mol -1 ] Freezing Point of water =273 K Atomic masses: C: 12.0 u , O: 16.0 u , H: 1.0 u ]

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

JEE MainJEE Main 2021Solutions

Solution:

$k _{ f }=1.86 \,k . kg / mol$
$T _{ f }^{\circ}=273 \,k$
solvent $: H _{2} O (625\, g )$
image
solute
$\Rightarrow \Delta T _{ f }= k _{ f } \times m$
$\Rightarrow \left( T _{ f }^{\circ}- T _{ f }^{1}\right)=1.86 \times \frac{83 / 62}{624 / 1000}$
$\Rightarrow 273- T _{ f }^{1}=\frac{1.86 \times 83 \times 1000}{62 \times 625}$
$=\frac{154380}{38750}$
$\Rightarrow 273- T _{ f }^{1}=4$
$\Rightarrow T _{ f }^{1}=259 \,K$