Question

A 13% solution (by weight) of sulphuric acid with a density of $1.02g{ml}^{-1}$? To what volume should 100 mL of this acid be diluted in order to prepare a 1.5-N solution?

Answer 1

$Molarity=\frac{Percentagebyweight\times Density\times 10}{Molecualrweight}$
$=\frac{13\times 1.02\times 10}{98}=\frac{132.6}{98}=1.35M$
13 % solution by weight means that 13 g of solute is dissolved in 87 g of solvent.
Thus
, $molality=\frac{\frac{Weightofsolute}{Molecularweightofsolute}}{Weightofsolvent}\times 1000$

$=\frac{\frac{13}{98}}{87}\times 1000$
$=\frac{13\times 1000}{98\times 87}=1.52m$
$Normality=Molarity\times Ewfactor$
$\therefore N=1.35\times 2=2.70N$
For dilution
$N_1{V}_1=N_2{V}_2$
$100\times 2.70N=1.5N\times V_2$
$V_2=180mL$
So, the acid should be diluted upto 180 mL to prepare 1.5 N solution.