A 0.2 g sample of chromite was fused with excess of N a 2 O 2 and brought into solution according to reaction: 2Fe ( Cr O 2 ) 2 + 7N a 2 O 2 → 2NaFe O 2 + 4N a 2 Cr O 4 + 2N a 2 O

Question

A 0.2 g sample of chromite was fused with excess of $N a_{2} O_{2}$ and brought into solution according to reaction: $2Fe (Cr O_{2})_{2}$ + $7N a_{2} O_{2}$ $\to$ $2NaFe O_{2}$ + $4N a_{2} Cr O_{4}$ + $2N a_{2} O$ . The solution was acidified with dil. $HCl$ and 3.92 g Mohr's salt (molar mass = 392 g/mol) was added. The excess of $F e^{2 +}$ required 80 ml of 0.05 N - $K_{2} C r_{2} O_{7}$ for titration. What is the per cent of Cr in sample? (Cr = 52, Fe = 56)

Answer:

52.00

Solution:

Mohr's salt = 3.92/392 = 10 mmol = 10 meq. Excess $F e^{2 +}$ = $80 \times 0.05$ = 4 meq. Consumed by chromate = 6 meq = eq of Cr. Mass Cr = 6×52/3 = 104 mg = 0.104 g. %Cr = 0.104/0.2×100 = 52%. Ans 0052.

TestHub · Accepted Answer

Mohr's salt = 3.92/392 = 10 mmol = 10 meq. Excess F e 2 + = 80 × 0.05 = 4 meq. Consumed by chromate = 6 meq = eq of Cr. Mass Cr = 6×52/3 = 104 mg = 0.104 g. %Cr = 0.104/0.2×100 = 52%. Ans 0052.

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