When $35 mL$ of $0.15M$ lead nitrate solution is mixed with $20 mL$ of $0.12M$ chromic sulphate solution, _____ $\times {10}^{-5}$ moles of lead sulphate precipitate out.(Round off to the Nearest integer).

$$\begin{gathered} 3 Pb{\left({NO}_3\right)}_2+{Cr}_2{\left({SO}_4\right)}_3\to 3{PbSO}_4+2Cr{\left({NO}_3\right)}_3 \\ 35ml,0.15M 20ml,0.12M \\ 5.25m.mol 2.4m.mol \end{gathered}$$

[Since, n=MV(l)=MV(ml) $\times$10^-3]

1mole of $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$  ------ 3 moles of ${PbSO}_4$

5.25moles of $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ ----        ?

[Since $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ is the limiting reagent]

moles of ${PbSO}_4$ formed $=5.25\times {10}^{-3}$

$=525\times {10}^{-5}$.

Hence the correct answer is 525.