Question

In an ore the only oxidisable material is $S{n}^{2+}$. This ore is titrated with a dichromate solution containing $2.5g{K}_{2}C{r}_2{O}_7$ in $0.50litre$. A $0.40g$ of sample of the ore required $10.0c{m}^3$ of the titrant to reach equivalent point. Calculate the percentage of tin in ore. $(K=39.1,Cr=52,Sn=118.7)$.

Answer 1

Mol. mass of $K_{2}C{r}_2{O}_7=2\times 39.1+2\times 52+7\times 16$
$=78.2+104.0+112.0$
$=294.2$
Eq. mass of $K_{2}C{r}_2{O}_7=\frac{294.2}{6}=49.03$
Normality of $K_{2}C{r}_2{O}_7$ solution $=\frac{2.5}{49.03}\times \frac{1000}{500}=\frac{5}{49.03}N$
$10mL\frac{5}{49.03}N{K}_{2}C{r}_2{O}_7\equiv 10mL\frac{5}{49.02}N$ stannous ion
Eq. mass of $S{n}^{2+}=\frac{118.7}{2}=59.35$
Amount of $Sn$ in the sample $=\frac{5}{49.03}\times \frac{59.35}{1000}\times 10$
$=0.0605g$
Percentage of $Sn$ in the ore $=\frac{0.0605}{0.40}\times 100=15$.