Some Basic Concepts of Chemistry - Result Question 63
####1. 5 moles of $A B _2$ weight $125 \times 10^{-3} kg$ and 10 moles of $A _2 B _2$ weight $300 \times 10^{-3} kg$. The molar mass of $A\left(M _A\right)$ and molar mass of $B\left(M _B\right)$ in $kg mol^{-1}$ are
(a) $M _A=10 \times 10^{-3}$ and $M _B=5 \times 10^{-3}$
(2019 Main, 12 April I)
(b) $M _A=50 \times 10^{-3}$ and $M _B=25 \times 10^{-3}$
(c) $M _A=25 \times 10^{-3}$ and $M _B=50 \times 10^{-3}$
(d) $M _A=5 \times 10^{-3}$ and $M _B=10 \times 10^{-3}$
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Solution:
Key Idea To find the mass of $A$ and $B$ in the given question, mole concept is used.
Number of moles $(n)=\frac{\text { given mass }(w)}{\text { molecular mass }(M)}$
| Compound | Mass of $A(g)$ | Mass of $B(g)$ |
|---|---|---|
| $A B _2$ | $M _A$ | $2 M _B$ |
| $A _2 B _2$ | $2 M _A$ | $2 M _B$ |
We know that,
Number of moles $(n)=\frac{\text { given mass }(w)}{\text { molecular mass }(M)}$
$$ n \times M=w $$
Using equation (A), it can be concluded that
$$ \begin{aligned} 5\left(M _A+2 M _B\right) & =125 \times 10^{-3} kg \ 10\left(2 M _A+2 M _B\right) & =300 \times 10^{-3} kg \end{aligned} $$
From equation (i) and (ii)
$$ \frac{1}{2} \frac{\left(M _A+2 M _B\right)}{\left(2 M _A+2 M _B\right)}=\left(\frac{125}{300}\right) $$
On solving the equation, we obtain
$$ \text { and } \quad \begin{aligned} & M _A=5 \times 10^{-3} \ & M _B=10 \times 10^{-3} \end{aligned} $$
So, the molar mass of $A\left(M _A\right)$ is $5 \times 10^{-3} kg mol^{-1}$ and $B\left(M _B\right)$ is $10 \times 10^{-3} kg mol^{-1}$.
