When $35 \mathrm{~mL}$ of $0.15 \mathrm{M}$ lead nitrate solution is mixed with $20 \mathrm{~mL}$ of $0.12 \mathrm{M}$ chromic sulphate solution, _______________ $\times 10^{-5}$ moles of lead sulphate precipitate out. (Round off to the Nearest
(525)
$3 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3} \rightarrow 3 \mathrm{PbSO}_{4}+2 \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}$
$35 \mathrm{ml} \quad 20 \mathrm{ml}$
$0.15 \mathrm{M} \quad 0.12 \mathrm{M}$
$=5.25 \mathrm{~m} . \mathrm{mol}=2.4 \mathrm{~m} \cdot \mathrm{mol} \quad 5.25 \mathrm{~m} \cdot \mathrm{mol}$
$=5.25 \times 10^{-3} \mathrm{~mol}$
therefore moles of $\mathrm{PbSO}_{4}$ formed $=5.25 \times 10^{-3}=525 \times 10^{-5}$
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