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AIIMS 2006
Exam
AIIMS
Year
2006
Subject
Chemistry
Chapter
Solutions

AIIMS 2006 · Chemistry · Solutions

A 5 \% solution (by mass) of cane sugar in water has freezing point...

A 5%5 \% solution (by mass) of cane sugar in water has freezing point of 271 K271 \mathrm{~K} and freezing point of pure water is 273.15 K273.15 \mathrm{~K}. The freezing point of a 5%5 \% solution (by mass) of glucose in water is

Options

  1. A271 K271 \mathrm{~K}
  2. B273.15 K273.15 \mathrm{~K}
  3. C269.07 K269.07 \mathrm{~K}
  4. D277.23 K277.23 \mathrm{~K}.

Correct answer

C. 269.07 K269.07 \mathrm{~K}

Step-by-step solution

: K_fforwaterfor water=\frac{\Delta T_f \times W \times m}{1000 \times w}(where(whereW=wt.ofwater,wt. of water,w=wt.ofcanesugar,wt. of cane sugar,m=molecularwt.ofcanesugar)=molecular wt. of cane sugar) =\frac{2.15 \times 100 \times 342}{1000 \times 5}=14.7Now,for14.7 Now, for5 \%glucose,glucose,\begin{aligned}& \DeltaT_f=\frac{K_f \times 1000 \times w^{\prime}}{W \times m^{\prime}}\left(\text{ where })w^{\prime}=\right.\text{ wt. of glucose, }\\ &m^{\prime}=\text{ molecular wt. of glucose }\right)\\ &=\frac{14.7 \times 1000 \times 5}{100 \times 180}=4.08 \end{aligned}∴\therefore Freezing point of glucose solution =273.15−4.08=269.07 K=273.15-4.08=269.07 \mathrm{~K}.
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