Question

The mass of potassium dichromate crystals required to oxidise 750 cm${}^3$ of 0.6 M Mohr's salt solution is:

[Given, molar mass, potassium dichromate = 294, Mohr's salt = 392]

• A. 0.39g
• B. 0.37g
• C. 22.05g
• D. 2.2 g

Answer 1

The correct option is C 22.05g

Amount of Mohr's salt oxidized is $0.75mL\times 0.6mol/l=0.45mol$
$K_{2}C{r}_2{O}_7+6FeS{O}_4+7H_{2}S{O}_4\to K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+3F{e}_2(S{O}_4{)}_3+7H_{2}O$

For 6 mol of Mohr's salt 1 mol of Pottasium dichromate is required

For 0.45 mol of Mohr's salt,Pottasium dichromate required is $\frac{1\times 0.45}{6}=0.075mol$

Mass of potassium dichromate required is $0.075\times 294=22.05g$

Hence, the correct option is $C$