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Lewis Kossel Structure

Calculate the...

Question

Calculate the mass of ascorbic acid $(V i t a m i n C, C_{6} H_{8} O_{6})$ to be disssolved in 75 g of acetic acid to lower its melting point by $1.5. K_{f} = 3.9 K^{- 1}$ .

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Solution

We have,

Mass of acetic acid, $W_{1} = 75 g$

Molar mass of the ascorbic acid

$(C_{6} H_{8} O_{6}),$

$M_{2} = 6 \times 12 + 1 \times 8 + 6 \times 16$

$= 176 g / m o l$

So, lowering of melting point, $Δ t f = 1.5 k$

By using formula:

$M_{2} = \frac{1000 k f W_{2}}{W_{1} Δ t f}$

$W_{2} = \frac{M_{2} \times W_{1} \times Δ t f}{1000 \times k f}$

$= \frac{(176 g / m o l) (75 g) (1.5 k)}{(1000 g k g^{- 1}) (3.9 k k g m o l^{- 1})}$

$W_{2} = 5.08 g$

Thus the required mass in ascorbic acid is $5.08 g$

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C_{6} H_{8} O_{6}

1 {.5}^{\circ} C . K_{f} = 3.9 K kg mo l^{- 1}

(C_{6} H_{8} O_{6})

75

{1.5}^{0} C

3.9

{m o l}^{- 1}

(Vitamin C, C_{6} H_{8} O_{6}

75 g

{1.5}^{\circ} C . K_{f} = 3.9 K K g m o l^{- 1}

0.02

277.4 K

C_{6} H_{6} = 278.4 K

K_{f}

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= 5.12 K m o l^{- 1} k g)

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