Some Basic Concepts of Chemistry 1 Question 1
1. 5 moles of $A B_{2}$ weight $125 \times 10^{-3} \mathrm{~kg}$ and 10 moles of $A_{2} B_{2}$ weight $300 \times 10^{-3} \mathrm{~kg}$. The molar mass of $A\left(M_{A}\right)$ and molar mass of $B\left(M_{B}\right)$ in $\mathrm{kg} \mathrm{mol}^{-1}$ 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
$$ \begin{aligned} 5\left(M_{A}+2 M_{B}\right) & =125 \times 10^{-3} \mathrm{~kg} \ 10\left(2 M_{A}+2 M_{B}\right) & =300 \times 10^{-3} \mathrm{~kg} \end{aligned} $$
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 $5 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}$ and $B\left(M_{B}\right)$ is $10 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}$.
