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Q.The boiling point of benzene is 353.23 K. On dissolving 1.80 g non-volatile solute in benzene's 90 g, the boiling point rises to 354.11 K. Find out the molar mass of the solute. For Benzene, Kb=2.53 K Kg mol−1K_b = 2.53\ K\,Kg\,mol^{-1}.

(OR)
Explain the following -
(a) Osmotic Pressure
(b) Raoult's Law
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2024Subjective· 2mImportance★★★★★
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Using ΔTb=Kb×m\Delta T_b = K_b \times m, the molar mass of the solute works out to about 57.5 g/mol.

Given: ΔTb=354.11−353.23=0.88 K\Delta T_b = 354.11 - 353.23 = 0.88\ K, mass of solute w2=1.80 gw_2 = 1.80\ g, mass of solvent (benzene) w1=90 g=0.090 kgw_1 = 90\ g = 0.090\ kg, Kb=2.53 K kg mol−1K_b = 2.53\ K\,kg\,mol^{-1}.

Elevation in boiling point is related to molality by ΔTb=Kb×m\Delta T_b = K_b \times m, so:

m=ΔTbKb=0.882.53=0.3478 mol kg−1m = \frac{\Delta T_b}{K_b} = \frac{0.88}{2.53} = 0.3478\ mol\,kg^{-1} …

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