When 35 mL of 0.15 M lead nitrate solution

$3 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3} \rightarrow 3 \mathrm{PbSO}_{4}+2 \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}$

$\begin{array}{ll}35 \mathrm{ml} & 20 \mathrm{ml} \\ 0.15 \mathrm{M} & 0.12 \mathrm{M}\end{array}$

$=5.25 \mathrm{~m} \cdot \mathrm{mol}=2.4 \mathrm{~m} \cdot \mathrm{mol} 5.25 \mathrm{~m} \cdot \mathrm{mol}$

$=5.25 \times 10^{-3} \mathrm{~mol}$

therefore moles of $\mathrm{PbSO}_{4}$ formed $=5.25 \times 10^{-3}$

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