Question

250 mL of a solution contains 6.3g of oxalic acid (mol. wt. =126). What is the volume (in litres) of water to be added to this solution to make it a 0.1 solution?

• A. 750
• B. 7.5
• C. 0.075
• D. 0.75

Answer 1

Answer: (D)

$\because$ 250 mL of solution contains oxalic acid = 6.3g $\therefore$ 1000 mL (1 L) of solution contains oxalic $acid=\frac{6.3}{250}\times 1000$ $=25.2g/L$ or Strength = 25.2 / L $S=E\times N$ where, S = strength E = Equivalent wt. N = Normality $\text{Eq}\text{.}\text{wt}\text{. = }\frac{\text{Molecular}\text{wt}\text{.}}{\text{basicity}\text{of}\text{acid}}$ Molecular wt. of oxalic acid = 126 Basicity =2 $=\frac{126}{2}=63$ So, $S=E\times N$ $25.2=63\times N$ $N=\frac{25.2}{63}=0.4$ Hence, the given solution is 250 mL 0.4 N, oxalic acid solution. Now, this solution is makes to 0.1 N solution as: $N_1{V}_1=N_2{V}_2$ $0.4\times 250=0.1\times V_2$ $V_2=\frac{0.4\times 250}{0.1}=1000mL$ So, water added = 1000 - 250 = 750 mL water = 0.75 L water