10 g of MnO 2 on reaction with conc HCl liberated 0.1 equivalent of Cl 2( Mn =55). Hence, per cent purity of MnO 2 is

Question

10 g of MnO 2 on reaction with conc HCl liberated 0.1 equivalent of Cl 2( Mn =55). Hence, per cent purity of MnO 2 is (A) 87.0 (B) 21.75 (C) 50.0 (D) 43.5

Tardigrade Admin · Accepted Answer

MnO 2+ HCl arrow MnCl 2+ Cl 2+2 H 2 O 0.1 equivalent Cl 2=(0.1/2) mol Cl 2- =0.05 mol MnO 2 =4.35 g MnO 2 in 10 g sample Thus, 43.5 % pure

Q. $10 g$ of $M n O_{2}$ on reaction with conc $H Cl$ liberated $0.1$ equivalent of $C l_{2} (M n = 55)$ . Hence, per cent purity of $M n O_{2}$ is

$87.0$

$21.75$

$50.0$

$43.5$

$M n O_{2} + H Cl \to M n C l_{2} + C l_{2} + 2 H_{2} O$
$0.1$ equivalent $C l_{2} = \frac{0.1}{2} m o lC l_{2}^{-}$
$= 0.05 m o l M n O_{2}$
$= 4.35 g M n O_{2}$ in $10$ g sample
Thus, $43.5%$ pure

Q. 10g of MnO2​ on reaction with conc HCl liberated 0.1 equivalent of Cl2​(Mn=55). Hence, per cent purity of MnO2​ is

87.0

21.75

50.0

43.5