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

We are given 35 mL of 0.15 M lead nitrate ($$Pb(NO_3)_2$$) and 20 mL of 0.12 M chromic sulphate ($$Cr_2(SO_4)_3$$). The precipitation reaction is: $$Pb^{2+} + SO_4^{2-} \rightarrow PbSO_4 \downarrow$$.

Moles of $$Pb^{2+}$$ ions = $$0.035 \times 0.15 = 0.00525$$ mol.

Since each formula unit of $$Cr_2(SO_4)_3$$ provides 3 sulphate ions, the moles of $$SO_4^{2-}$$ ions = $$0.020 \times 0.12 \times 3 = 0.0072$$ mol.

The reaction between $$Pb^{2+}$$ and $$SO_4^{2-}$$ is in a 1:1 molar ratio. Since 0.00525 mol of $$Pb^{2+}$$ is less than 0.0072 mol of $$SO_4^{2-}$$, lead is the limiting reagent.

Therefore, moles of $$PbSO_4$$ precipitated = $$0.00525$$ mol = $$525 \times 10^{-5}$$ mol.

The answer is $$525$$.