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

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.