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Q.The boiling point of benzene is 353.23K. When 1.80g of non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11K. Calculate the molar mass of solute. (Kb for Benzene is 2.53 K kg mol⁻¹)

Punjab PsebPSEB Punjab Class 12 Board 2020Subjective· 3mImportance★★★★★
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Using the elevation-in-boiling-point formula M2=1000 Kb w2ΔTb w1M_2=\dfrac{1000\,K_b\,w_2}{\Delta T_b\,w_1} gives molar mass ≈57.5 g/mol\approx 57.5\ g/mol.

Elevation in boiling point:

ΔTb=354.11−353.23=0.88 K\Delta T_b = 354.11 - 353.23 = 0.88\ K

The formula relating elevation in boiling point to molar mass of solute is:

ΔTb=Kb m=Kb×w2×1000M2×w1\Delta T_b = K_b\,m = K_b \times \dfrac{w_2 \times 1000}{M_2 \times w_1}

where w2=1.80 gw_2 = 1.80\ g (mass of solute), w1=90 gw_1 = 90\ g (mass of solvent, benzene), Kb=2.53 K kg mol−1K_b = 2.53\ K\,kg\,mol^{-1}.

Rearranging for molar mass M2M_2: …

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