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Question
Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.
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Solution
Given: Lowering in melting point (ΔTf) = 1.5°C
Mass of solvent (CH3COOH), w1 = 75 g
Mass of solute, w2 = ?
Molar mass of solvent (CH3COOH), M1 = 60 g mol−1
Molar mass of solute (C6H8O6), M2 = 6 × 12 + 8 × 1 + 6 × 16
= 72 + 8 + 96
= 176 g mol−1
Kf = 3.9 K kg mol−1
Applying the formula,
ΔTf = `(K_f xx w_2 xx 1000)/(M_2 xx w_1)`
⇒ w2 = `(Delta T_f xx M_2 xx w_1)/(K_f xx 1000)`
= `(1.5 xx 176 xx 75)/(3.9 xx 1000)`
= 5.08 g (approx)
Hence, 5.08 g of ascorbic acid needs to be dissolved.
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