Verified Solution • 2025–26 Session moderate

QJEEM21Ch9242: JEE Main 2021 Chemistry - Class 12 JEEM21Ch9242 Solution: 83 g of ethylene glycol dissolved in 625 g of water. The…

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]

Step-by-Step Verified Solution

kf = 1.86 kg/mol

T fo_f^o = 273 K

solvent : H2O(625 g)

Solute : 83 g of ethylene glycol

⇒\Rightarrow Δ\Delta Tf = kf ×\times m

⇒\Rightarrow (Tfo−Tf1)=1.86×83/62625/1000\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}

⇒273−Tf1=1.86×83×100062×625=15438038750\Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}

⇒273−Tf1=4\Rightarrow 273 - T_f^1 = 4

⇒Tf1=259\Rightarrow T_f^1 = 259 K

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