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Chemistry Question on Colligative Properties

At 300 K,36 g300 \,K , 36\, g of glucose present per litre in its solution has an osmotic pressure of 4.984.98 bar. If the osmotic pressure of solution is 1.521.52 bar at the same temperature, what would be its concentration?

A

11 g L−111 \,g\, L^{-1}

B

22 g L−122 \,g\,L^{-1}

C

36 g L−136 \,g\,L^{-1}

D

42 g L−142\, g\,L^{-1}

Answer

11 g L−111 \,g\, L^{-1}

Explanation

Solution

4.98=36×S×300180×1 4.98 =\frac{36 \times S \times 300}{180 \times 1} S=0.083S =0.083 bar L mol−1K−1L\, mol ^{-1} K ^{-1} π2=n2ST2V2\pi_{2} =\frac{n_{2} S T_{2}}{V_{2}} 1.52=n2×0.083×300V2 1.52 = \frac{n_2 \times 0.083 \times 300}{V_2} n2V2=0.061 mol L−1=11 g L−1\frac{n_2}{V_2} = 0.061\,mol\,L^{-1} = 11\,g\,L^{-1}