Question

What volume of decinormal $K_2{Cr}_2{O}_7$ solution would be needed to convert 1.19 g of tin to stannic chloride in an acidic medium? ($A_{Sn}=119$)

• A. 400 ml
• B. 200 ml
• C. 250 ml
• D. 300 ml

Answer 1

The correct option is A 400 ml

No.of moles of tin in $1.19g$ of tin$=\frac{1.19}{119}$

$n=0.01$

as $Sn$ is converting to $Sn{Cl}_4$ i.e. ${Sn}^0\to {Sn}^{+4}$

Thus $nf=4$

Normality of ${Kr}_2{Cr}_2{O}_7=O.1N$

$\therefore 0.1N\times V=0.01\times 4$

$V=0.42=400ml$

$400ml$ of volume of decinormal ${Kr}_2{Cr}_2{O}_7$ solution would be needed.