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

Open in App

Solution

Given:
$Δ T_{f} = {1.5}^{\circ} C o r K, K_{f} = 3.9 K k g m o l^{- 1}$
Mass of acetic acid $(w_{A}) = 75 g$
Molar mass of ascorbic acid $(M_{B}) = 176 g m o l^{- 1}$
Mass of ascorbic acid

We know that, $Δ T_{f} = K_{f} \times m o l a l i t y (m)$
We also know, molality(m)= $\frac{mass of solute(B)}{molar mass of solute(B) \times mass of solvent(A) (i n k g)}$

So, $Δ T_{f} = \frac{K_{f} \times w_{B}}{M_{B} \times w_{A}}$

$\Rightarrow 1.5 K = \frac{3.9 K K g m o l^{- 1} \times W_{B}}{176 g m o l^{- 1} \times 75 \times 10^{- 3} k g}$

$\Rightarrow W_{B} = \frac{1.5 \times 75 \times 176 \times 10^{- 3}}{3.9} = 5.077 g$

Hence, required mass of ascorbic acid is $5.077 g$ .

Suggest Corrections

Similar questions

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

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

(V i t a m i n C, C_{6} H_{8} O_{6})

1.5. K_{f} = 3.9 K^{- 1}

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

75

{1.5}^{0} C

3.9

{m o l}^{- 1}

256 g {m o l}^{- 1}

75 g

0.48 K

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

Acetic acid associates in benzene. 160 g of acetic acid when dissolved in 100 g

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program