Chemistry - Concentration Terms Question with Solution | TestHub

$3\mathrm{NaOH}+{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{Na}}_3{\mathrm{PO}}_4+3{\mathrm{H}}_2\mathrm{O}$

required $0.49\mathrm{gm}$

$$\begin{array}{rlr}\text{ moles }=3\times 0.005& =\frac{0.49}{98}\text{ moles }=0.015& =0.005\text{ moles }\end{array}$$

moles of $\mathrm{NaOH}={\mathrm{M}}_{\mathrm{NaOH}}\times {\mathrm{V}}_{\mathrm{NaOH}}$ (litre)

$$\begin{array}{rlr}& 0.015=0.2\times {\mathrm{V}}_{\mathrm{NaOH}}& {\mathrm{V}}_{\mathrm{NaOH}}=0.0751=75\mathrm{ml}\end{array}$$