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Question

Chemistry Question on Solutions

At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of a urea solution is 1.52 bar its at the same temperature, concentration is

A

1.6 M

B

0.61 M

C

0.061 M

D

None of these

Answer

0.061 M

Explanation

Solution

π1V=n1RT\pi_1 V = n_1 RT
π2V=n2RT\pi_2 V = n_2RT
π1π2=n1n2\frac{\pi_1}{\pi_2} = \frac{n_1}{n_2}

n1=36180=15n_1=\frac {36}{180}=\frac 15 mol
π1\pi_1 = 4.98 bar
π2\pi_2 = 1.52 bar
4.981.52=15n2\frac{4.98}{1.52} = \frac{\frac 15}{n_2}

n2=1.524.98×5n_2 = \frac{1.52}{4.98 \times 5} = 0.061 M
Osmotic pressure is a colligative property

\therefore 0.061 M Glucose = 0.061 M urea