Question

82. 100 mL of H2SO4 solution having molarity 1 M and density 1.5 g/mL is mixed with 400 mL of
water. Calculate final molarity of H2SO4 solution, if final density is 1.25 g/mL:
(a) 4.4 M
(b) 0.145 M
(c) 0.52 M
(d) 0.227 M
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Answer 1

We know that volumes of various solutions are not additive in nature.

Therefore, we can infer that 100 $ml$ of Hydrogen Sulphate and 400 $ml$ of water do not actually make 500 $ml$ of the resulting solution but instead create a solution which has a lesser amount than that.

Yet, we have 150 g of acid 1M + 400g of water, which is equivalent to 550 g of material as well as a final density of 1.25 $g/ml$ . Then;

$V = \frac{M}{d} =$ $\frac{550g}{1.25g/ml} = 440 ml$

Initially, we had;

$n = \frac{V}{M} = \frac{0.1 l}{1 mol/l} = 0.1 mol$

The amount derived is that of $H_2SO_4$.

Therefore;

$M = (\frac{mol}{V}) = \frac{0.1 mol}{0.44l} = 0.227 mol/l = 0.227 M$

Ans) (d) 0.227 M