Calculate the mass of ascorbic acid (Vitamin C. $C_{6} H_{8} O_{6}$ ) to be dissolved in 75 g of acetic acid to lower its melting point by $1 {.5}^{\circ} C . K_{f} = 3.9 K kg mo l^{- 1}$ .

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Solution

Mass of acetic acid, w1= 75 g Molar mass of ascorbic acid ( $C_{6} H_{8} O_{6}$ ),

$M 2 = 6 \times 12 + 8 \times 1 + 6 \times 16$

$= 176 g m o l^{- 1}$

Lowering of melting point, $Δ T_{f} = 1.5 K$

We know that:
$Δ T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$

$w_{2} = \frac{Δ T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000}$

$w_{2} = \frac{1.5 \times 176 \times 75}{3.9 \times 1000}$

$w_{2} = 5.08 g$
Hence, 5.08 g of ascorbic acid is needed to be dissolved.

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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=3.9Kkgmol−1.

Mass of acetic acid, w1= 75 g Molar mass of ascorbic acid (C6H8O6),M2=6×12+8×1+6×16=176gmol−1Lowering of melting point, ΔTf=1.5KWe know that:ΔTb=Kb×1000×w2M2×w1w2=ΔTb×M2×w1Kb×1000w2=1.5×176×753.9×1000w2=5.08gHence, 5.08 g of ascorbic acid is needed to be dissolved.

Calculate the mass of ascorbic acid (Vitamin C. $C_{6} H_{8} O_{6}$ ) to be dissolved in 75 g of acetic acid to lower its melting point by $1 {.5}^{\circ} C . K_{f} = 3.9 K kg mo l^{- 1}$ .