Question

Calculate the $\%$ of $Cr$ in a sample of dichromate ore:

• A. $15.22$
• B. $17.22$
• C. $16.22$
• D. $18.22$

Answer 1

Answer: (A)

$15.22$

g eq of ferrous ammonium sulphate $=\frac{50}{1000}\times 0.12=0.006$ g eq

Also 1 ml dichromate is equal to $0.006g$ of $Fe$.

$15.05$ mL of dichromate will be equal to $15.05\times 0.006=0.0903g$ of $Fe$.

Equivalent mass of $Fe$ is $55.8$ g/eq

Number of g eq of $Fe=\frac{0.0903g}{55.8g/eq}=0.0016183eq$

Out of $0.006$ g eq of ferrous ammonium sulphate, $0.0016183$ g eq is excess.

Number of g eq of ferrous ammonium sulphate that reacts with ore sample

$=0.006-0.0016183=0.004382geq$.

Equivalent mass of $K_{2}C{r}_2{O}_7=49g/eq$

Mass of $K_{2}C{r}_2{O}_7=49\times 0.004382=0.2147$

Molar mass of $K_{2}C{r}_2{O}_7=294g/mol$

Mass of $Cr=2\times \frac{52.0}{294}\times 0.2147=0.0759g$

Percent of $Cr$ in the ore $=\frac{0.0759\times 100}{0.5}=15.22$%

Hence, the correct option is $\text{A}$