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A 5% solution(by mass) of cane sugar in water has freezing point of 271k and freezing point of pure water is 273.15k. The freezing point of a 5% solution (by mass) of glucose in water is

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271k

273.15k

269.07k

277.23k

answer is B.

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Detailed Solution

FP of pure water = 273.15K

FP of 5% (W/W) C₁₂H₂₂O₁₁(aq.) = 271K

FP of 5% (W/W) C₆H₁₂O₆(aq.) = ?

$Delta T_f;of;C_{12}H_{22}O_{11}=(273.15-273)=2.15K$

$Delta T_f=K_fm$

$because$ the solvent is water in both cases

By comparing Sucrose and glucose solutions;

$frac{Delta T_1}{Delta T_2}=frac{m_1}{m_2}$

$frac{Delta T_{sucrose}}{Delta T_{glucose}}=frac{m_{sucrose}}{m_{glucose}}$

both are 5% by mass

$frac{2.15}{Delta T_{glucose}}=frac{(frac{5}{342}times frac{1000}{95})}{(frac{5}{180}times frac{1000}{95})}$

$frac{2.15}{Delta T_{glucose}}=frac{180}{342}$

$Delta T_{glucose}=frac{342times 2.15}{180}=4.085K$

$Delta T_f=273.15-T_f$

T_f = FP of glucose sol. =273.15 - 4.085 =269.07K

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