JEE MAIN - Chemistry (2021 - 27th August Evening Shift - No. 17)
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is __________ K. (Nearest integer) [Given : Kf = 1.86 K kg mol$$-$$1; Density of water = 1.00 g cm$$-$$3; Freezing point of water = 273.15 K]
Answer
271
Explanation
Molality = $${{\left( {{{40} \over {180}}} \right)mol} \over {0.2\,Kg}} = \left( {{{10} \over 9}} \right)$$ molal
$$ \Rightarrow \Delta {T_f} = {T_f} - {T_f}' = 1.86 \times {{10} \over 9}$$
$$ \Rightarrow {T_f}' = 273.15 - 1.86 \times {{10} \over 9}$$
= 271.08 K
$$ \simeq $$ 271 K (nearest - integer)
$$ \Rightarrow \Delta {T_f} = {T_f} - {T_f}' = 1.86 \times {{10} \over 9}$$
$$ \Rightarrow {T_f}' = 273.15 - 1.86 \times {{10} \over 9}$$
= 271.08 K
$$ \simeq $$ 271 K (nearest - integer)
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