Question

A 10 g mixture of glucose and urea present in 250 mL solution shows the osmotic pressure of 7.4 atm at ${27}^{o}C$.What is the percentage composition of mixture?

• A. Glucose $=$ 17.6%; Urea $=$ 82.4 %
• B. Glucose $=$ 19.6%; Urea $=$ 80.4 %
• C. Glucose $=$ 82.4%; Urea $=$ 17.6 %
• D. None of these

Answer 1

The correct option is C Glucose $=$ 82.4%; Urea $=$ 17.6 %

Let $x$ g glucose and $10-x$ g urea be present.

The molar masses of glucose and urea are $180$ g/mol and $60$ g/mol respectively.

The number of moles of glucose $=\frac{x}{180}$

The number of moles of urea $=\frac{10-x}{60}$

Total number of moles of solutes $=\frac{x}{180}+\frac{10-x}{60}=\frac{x+30-3x}{180}=\frac{30-2x}{180}=\frac{15-x}{90}$

The molarity of the solution is $C=\frac{15-x}{90}\times \frac{1000}{250}=\frac{30-2x}{45}$
The osmotic pressure $=\Pi =CRT$

Here, $R$ is the ideal gas constant and $T$ is the absolute temperature.

$7.4=\frac{30-2x}{45}\times 0.08206\times 300$

$30-2x=13.532x=16.47x=8.24$

$10-x=10-8.24=1.76$

Percentage of glucose $\frac{100x}{10}=10x=10\times 8.24=82.4\%$

Percentage of urea $=\frac{100(10-x)}{10}=10\left(1.76\right)=17.6\%$