Question

What is the molarity and molality of a $13\%$ solution (by weight) of sulphuric acid with a density of $1.02\, g/mL$ ? To what volume should $100\, mL$ of this solution be diluted in order to prepare a $1.5 \,N$ solution ?

Answer 1

Let us consider $1.0\, L$ of solution.
Weight of solution $= 1000 \times 1.02 = 1020 \,g$
Weight of $H_2SO_4 =1020 \times \frac{13}{100} = 132.60\, g$
Weight of $H_2O =1020-132.60=887.40\,g$
$\Rightarrow $ Molarity $=\frac{132.60}{98}=1.353\, M$
Molarity $=\frac{132.60}{98} \times \frac{1000}{887.40}=1.525 \,m$
Normality $=2 \times M =2.706 $
$\Rightarrow 2.706 \times 100 =1.5 \,V$
$\Rightarrow V=180.40\, mL$