Question

1.725 g of metal carbonate is mixed with 300 ml of N/10 hcl.
10 ml of N/2 sodium hydroxide were required to neutralize the excess of the acid calculate the equivalent mass of metal carbonate?​

Answer 1

Answer:

69

Explanation:

10mL of $\frac{N}{2}$ NaOH  solution

=10mL of $\frac{N}{2}$ HCl solution

≡50mL of$\frac{N}{10}$ HCl solution

Volume of $\frac{N}{10}$ HCl used for neutralisation =300-50=250 mL

250 mL of $\frac{N}{10}$ HCl≡250mL of$\frac{N}{10}$ metal carbonate solution

Let the equivalent mass of metal carbonate be E.

Mass of metal carbonate present in solution

=$\frac{N X E XV}{1000}$=1.725

=$\frac{1 X EX 250}{10 X 1000}$=1.725

=$\frac{E}{40}$=1.725

E= 40×1.725 = 69