JEE Advance - Chemistry (1993 - No. 2)

Upon mixing 45.0 ml. of 0.25 M lead nitrate solution with 25.0 ml. of 0.10M chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble.

Moles of PbSO4 formed = 0.0075 moles, [Pb(NO3)2] = 0.05357 M, [Cr2(SO4)3] = 0.0714 M

Moles of PbSO4 formed = 0.015 moles, [Pb(NO3)2] = 0.107 M, [Cr2(SO4)3] = 0.143 M

Moles of PbSO4 formed = 0.0075 moles, [Pb(NO3)2] = 0.05357 M, [Cr2(SO4)3] = 0.0714 M, [NO3-] = 0.3214 M

Moles of PbSO4 formed = 0.0075 moles, [Pb(NO3)2] = 0.054 M, [Cr2(SO4)3] = 0.071 M

Moles of PbSO4 formed = 0.0075 moles, [Pb2+] = 0.05357 M, [NO3-] = 0.3214 M, [Cr3+] = 0.0714 M

Giải thích

The reaction is $$\to$$

IIT-JEE 1993 Chemistry - Some Basic Concepts of Chemistry Question 30 English Explanation

$$C{r_2}{(S{O_4})_3}$$ is limiting reagent as $${{2.5} \over 1} < {{11.25} \over 3}$$.

$$\therefore$$ $$2.5 - x = 0$$

$$ \Rightarrow x = 2.5$$ milimoles

$$\therefore$$ Moles of $$PbS{O_4}$$ formed $$ = 3x = 3 \times 2.5 = 7.5 \times {10^{ - 3}}$$ moles

After $$PbS{O_4}$$ precipitate formation in the solution $$Pb{(N{O_3})_2}$$ and $$Cr{(N{O_3})_2}$$ remains.

$$\therefore$$ Milimoles of remaining $$Pb{(N{O_3})_2}$$ is $$ = 11.25 - 3x = 11.25 - 3 \times 2.5 = 3.75$$

And milimoles of remaining $$Cr{(N{O_3})_2}$$ is $$ = 2x = 2 \times 2.5 = 5$$

$$\therefore$$ Molar concentration or molarity of $$Pb{(N{O_3})_2} = {{3.75 \times {{10}^{ - 3}}} \over {{{45 + 25} \over {1000}}}} = 0.054\,M$$

And molarity of $$Cr{(N{O_3})_2} = {{5 \times {{10}^{ - 3}}} \over {{{70} \over {1000}}}} = 0.071\,M$$

JEE Advance - Chemistry (1993 - No. 2)

Giải thích

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