JEE MAIN - Chemistry (2021 - 26th August Evening Shift - No. 20)
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]
[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]
Answer
269
Explanation
kf = 1.86 kg/mol
T$$_f^o$$ = 273 K
solvent : H2O(625 g)
Solute : 83 g of ethylene glycol
$$\Rightarrow$$ $$\Delta$$Tf = kf $$\times$$ m
$$\Rightarrow$$ $$\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$$
$$ \Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}$$
$$ \Rightarrow 273 - T_f^1 = 4$$
$$ \Rightarrow T_f^1 = 259$$ K
T$$_f^o$$ = 273 K
solvent : H2O(625 g)
Solute : 83 g of ethylene glycol
$$\Rightarrow$$ $$\Delta$$Tf = kf $$\times$$ m
$$\Rightarrow$$ $$\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$$
$$ \Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}$$
$$ \Rightarrow 273 - T_f^1 = 4$$
$$ \Rightarrow T_f^1 = 259$$ K
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