Question

$4.9$ g of$ K_{2} Cr_{2} O_{7}$ is taken to prepare 0.1 L of the solution. 10 mL of this solution is further taken to oxidise $Sn^{2+}$ ion into $Sn^{4+}$ ion $Sn^{4+}$ so produced is used in second reaction to prepare $Fe^{3+}$ ion then the millimoles of $Fe^{3+}$ ion formed will be (assume all other components are in sufficient amount) [Molar mass of $ K_{2} Cr_{2} O_{7} =294$ g ].

• A. $5$
• B. $20$
• C. $10$
• D. none of these

Answer 1

Answer: (C)

$ K_{2} Cr_{2} O_{7} +Sn^{2+} \to Sn^{4+} +Cr^{3+} $
$Sn^{4+} Fe^{2+} \to Fe^{3+}$
Meq. of $Sn^{4+} = $ Meq of $K_{2} Cr_{2}O_{7}$
Meq. of $Sn^{4+} = $ Meq of $Fe^{3+}$
Meq. of $Fe^{3+} = $ meq of $K_{2} Cr_{2}O_{7}$
$N_{K_2Cr_2O_7}=\frac{4.9\times6}{294 \times01}=1$
millimol $ \times n- $factor $= 1\times10$
millimol $= 10$