Question

Find the molality of $H_{2}S{O}_4$ solution whose specific gravity is 1.99 g/mL and 49% by volume of $H_{2}S{O}_4$

Answer 1

$1.98g=$ Specific gravit

$1000ml\times 1.98=1980gram$

Solution weight-Water weight of solutions

1980-510=1480 $H_{2}S{O}_4$

Numbers of moles of $H_{2}S{O}_4$=Mass of $H_{2}S{o}_4$/Molar mass of $H_{2}S{O}_4$

$=\frac{1480}{98}=15.10mol$

Molality(m)=No of moles/mars of solution

$\frac{15.10}{510}=0.029moles$