Question

10g of $ {\text{Mn}}{{\text{O}}_{\text{2}}} $ on reaction with $ {\text{HCl}} $ forms $ 2.24{\text{ L}} $ of $ {\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right) $ at NTP, the percentage impurity of $ {\text{Mn}}{{\text{O}}_{\text{2}}} $ :
 $ {\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + 4HCl}} \to {\text{MnC}}{{\text{l}}_{\text{2}}}{\text{ + C}}{{\text{l}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} $
(A) 87%
(B) 25%
(C) $ 33.3\% $
(D) 13%

Answer 1

Hint: To answer this question, you should recall the formula for finding the number of moles of a gas when we are given the volume occupied by the gas. The volume occupied by one mole of a substance in the vapour phase is known as the molar volume of the substance and it is equal to $ 22400{\text{ ml}} $ at STP. We shall calculate the moles of chlorine and thus, manganese dioxide formed and its mass. Then, we can find the amount of impurities in the mass given.

Formula used:
$ n = \dfrac{w}{M} = \dfrac{V}{{{V_M}}} $
Where, $ n $ represents the number of moles of the given substance
 $ w $ represents the given mass of the substance
 $ M $ represents the molar mass of the given substance
 $ V $ represents the volume occupied by the substance in the vapor phase
And $ {V_M} $ represents the molar volume of the gas, i.e. the volume which is occupied by one mole of the gas at Standard Temperature and Pressure conditions.
Complete step by step solution:
In the question, we are given the reaction occurring as,
 $ {\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + 4HCl}} \to {\text{MnC}}{{\text{l}}_{\text{2}}}{\text{ + C}}{{\text{l}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} $
From the given reaction, we can conclude that one mole of chlorine is formed by one mole of $ {\text{Mn}}{{\text{O}}_{\text{2}}} $
We are given 10g of the impure sample and $ 2.24{\text{ L}} $ of $ {\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right) $ are formed.
The number of moles of chlorine formed are
 $ \dfrac{{2.24{\text{L}}}}{{22.4{\text{L}}}} = 0.1{\text{ mol}} $
So, we can say that $ 0.1 $ moles of manganese dioxide are formed.
The molar mass of manganese dioxide is known to be $ 87{\text{ g/mol}} $ . Thus, we have the amount of manganese dioxide reacts $ = 87 \times 0.1 = 8.7{\text{ g}} $
Thus, the weight of impurities in 10 grams of the mixture is $ 10 - 8.7 = 1.3{\text{g}} $
The correct answer is D.

Note: Mole is the basic unit for the measurement of the amount of a substance in the International System of units, commonly known as the SI Units. One mole of a substance is defined as the amount of the substance that contains particles equal to the Avogadro’s number.