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NCERT Class XII Chemistry
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Question : 52 of 53
Marks: +1, -0
Calculate the mass of ascorbic acid (vitamin C,C6H8O6\mathrm{C}, \mathrm{C}_6\mathrm{H}_8\mathrm{O}_6) to be dissolved in 75 g acetic acid to lower its melting point by 1.5°C, Kf = 3.9 K kg mol−1^{-1}.
Solution:  
Lowering in melting point (ΔTf)=1.5∘(\Delta T_{f})=1.5^{\circ}
Mass of solvent (CH3COOH),w1=75g,(\mathrm{CH}_{3} \mathrm{COOH}), w_{1}=75 \mathrm{g},
Mass of solute, w2=?w_{2}=?
Molar mass of solvent (CH3COOH),M1=60gmol−1(\mathrm{CH}_{3} \mathrm{COOH}), M_{1}=60 \mathrm{g} \mathrm{mol}^{-1}
Molar mass of solute, C6H8O6=72+8+96=176gmol−1\mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}=72+8+96=176 \mathrm{g} \mathrm{mol}^{-1}
(pplying the formula, ΔTf=Kf×w2×1000M2×w1\Delta T_{f}=K_{f} \times \frac{w_{2} \times 1000}{M_{2} \times w_{1}}
or w2=M2×w1×ΔTf1000×Kfw_{2}=\frac{M_{2} \times w_{1} \times \Delta T_{f}}{1000 \times K_{f}}
=176×75×1.51000×3.9=5.08g=\frac{176 \times 75 \times 1.5}{1000 \times 3.9}=5.08 \mathrm{g}
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