Question

Mixing of 50 mL of 0.25 M lead nitrate solution with 25 mL of 0.10 M chromic
sulphate solution causes the precipitation of lead sulphate. The molarity of
Pb2+ and Cr3+ ions in the solution respectively are
0.0667M, 0.0667M
O 0.1M, 0.2M
0.125M, 0.05M
0.05M, 0.125M​

Answer 1

Answer:

The correct structure of tri bromooctaoxide

1)

1.O=Br-Br - Br-o

2.o - Br - Br-Br =0

3.O=Br-Br

4.O=Br-Br-Br=0

Answer 2

Answer:

0.0667M and 0.0667M is the concentration of $Pb^+^2$ and $Cr^+^3$ ions.

Explanation:

Reaction

$3Pb(NO_3)_2 + Cr_2(SO_4)_3 {\longrightarrow} 3PbSO_4 + 2Cr(NO_3)_3$

3 moles of $Pb(NO_3)_3$ is used so

No of moles = $0.25* 0.05 = 0.0125 moles$

1 mole of $Cr_2(SO_4)_3$ is used

No of moles = $0.10 * 0.025 = 0.0025 moles$

So $Cr_2(SO_4)_3$ is the limiting agent

Lead sulphate formed will be 3 times the concentration of chromic sulphate as it is the limiting agent.

$PbSO_4 = 3* 0.0025=0.0075 moles$

Total volume of the solution is = (50ml + 25 ml) = 75ml = 0.075l

Concentration of

$Pb^+2 \\= \frac{0.0125- 0.0075}{0.0075} \\=0.0667 M$

depens on concentration of $PbSO_4$

Concentration of $Cr^+^3$

$= \frac{0.0025*2}{0.075} \\= 0.0667 M$

Depends on concentration of $Cr(NO_3)_3$