# Suppose that A and B form the compounds ${ B }_{ 2 }{ A }_{ 3 }$ and ${ B }_{ 2 }{ A }$ . If 0.05 mol of ${ B }_{ 2 }{ A }_{ 3 }$ weighs 9 g and 0.1 mole of ${ B }_{ 2 }{ A }$ weighs 10 g, the atomic weight of A and B respectively are:(A) 30 and 40(B) 40 and 30(C) 20 and 5(D) 15 and 20

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Hint: For any compound, its molar mass is calculated by adding the masses of particular elements in it (taking consideration of atomicity as well). Here in this question, make two equations out of the two compounds (${ B }_{ 2 }{ A }_{ 3 }$ ,${ B }_{ 2 }{ A }$ ) and then equate it.

Let us consider, ‘x’ as atomic weight of A and ‘y’ as atomic weight of B.
As it is already given in the question that 0.5 mol of ${ B }_{ 2 }{ A }_{ 3 }$ weighs 9 g.

Therefore 1$\left( 0.05\ \times \ 20 \right)$ mol of ${ B }_{ 2 }{ A }_{ 3 }$ will weigh $\left( 9\ \times \ 20 \right)$ g = 180 g.
Put ‘x’ and ‘y’ as masses of A and B respectively and then write the equation as given below:
${ 2x\ +\ 3y\ = }$ mass of 1 mol of ${ B }_{ 2 }{ A }_{ 3 }$ = 180 g

${ \Rightarrow \ 2x\ +\ 3y\ =\ 180 }$ (i).

Similarly it's given that 0.1 mole of ${ B }_{ 2 }{ A }$weighs 10 g.
Therefore 1$\left( 0.1\ \times \ 10 \right)$ mol of ${ B }_{ 2 }{ A }$ will weigh $\left( 10 \ \times \ 10 \right)$g = 100 g.

Put ‘x’ and ‘y’ as masses of A and B respectively and then write the equation as given below:
${ 2y\ +\ x\ = }$ mass of 1 mol of ${ B }_{ 2 }{ A }$= 100g

${ \Rightarrow 2y\ +\ x\ =\ 100 }$ (ii).

Equate both the equations (i) and (ii),
We will get x = 40 and y = 30.

Therefore from the above atomic masses of A and B are 40 and 30 respectively, which has been satisfied by Option (B).

The answer for this question is Option (B) 40 and 30.

Avogadro’s number: ${ 6.022\ \times \ { 10 }^{ 23 }}$ or${ 6.022\ \times \ { 10 }^{ 23 } }\ { mol }^{ -1 }$ .
${ number\ of\ moles\ =\ \frac { given\ mass }{ molar\ mass } \ =\ \frac { given\ volume\ at\ STP }{ molar\ mass\ at\ STP } \ =\ \frac { given\ mass }{ avogadro\ number } }$