Chemistry - Concentration Terms Question with Solution | TestHub

${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$

$\begin{array}{ccc}0.4\mathrm{mole}& 0.2\mathrm{mole}& -\\ 0.3\mathrm{mol}& -& 0.1\mathrm{mol}\end{array}$

$\mathrm{Molarity}=\frac{\mathrm{Number} \mathrm{of} \mathrm{moles}}{\mathrm{Volume} \mathrm{in} \mathrm{litres}}$

Molarity of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is $\frac{0.1}{4}=0.025\mathrm{M}$

$=25\times {10}^{-3}\mathrm{M}$

Molarity of the resulting product ${\mathrm{Na}}_2{\mathrm{SO}}_4$ in the solution is $25$ millimoles