Thank you for reporting, we will resolve it shortly
Q.
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?
EAMCETEAMCET 2004
Solution:
$ \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 $ $ =25.2\,g/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}=1000\,mL $ So, water added = 1000 - 250 = 750 mL water = 0.75 L water