Question

Suppose the elements A and B combine to form two
compound AB and A2B. When 0.1 mole of AB weighs 20 g
and 0.5 mole of A2B weighs 125 g. The atomic weight of
A and B respectively are​

Answer 1

Let the atomic weight weight of A be 'a g/mol' and B be 'b g/mol' respectively.

As we know that,

$\bf Moles \: (n) = \dfrac{Given \ mass \ (w)}{Molecular \ mass \ (M)}$

Thus,

$\bf \leadsto w = n \times M$

According to the question;

0.1 mole of $\rm AB$ weighs 20 g. So,

$\rm \implies 0.1 \times (a + b) = 20 \\ \\ \rm \implies a + b = \dfrac{20}{0.1} \\ \\ \rm \implies a + b = 200 \: \: \: \: \: \: \: \: \: \: \: ...(i)$

0.5 mole of $\rm A_2B$ weighs 125 g. So,

$\rm \implies 0.5 \times (2a + b) = 125 \\ \\ \rm \implies 2a + b = \dfrac{125}{0.5} \\ \\ \rm \implies 2a + b = 250 \: \: \: \: \: \: \: \: \: \: \: ...(ii)$

On substracting (i) from (ii), we get:

$\rm \leadsto 2a + b - (a + b) = 250 - 200 \\ \\ \rm \leadsto 2a - a + b - b = 50 \\ \\ \rm \leadsto a = 50 \: g /mol$

On substituting value of a in (i), we get:

$\rm \leadsto 50 + b = 200 \\ \\ \rm \leadsto b = 200 - 50 \\ \\ \rm \leadsto b = 150 \: g/mol$

$\therefore$

Atomic weight of A = 50 g/mol

Atomic weight of B = 150 g/mol

Answer 2

0.1(A + B) = 20

A + B = 20/0.1

A + B = 200 (1)

0.5(2A + B) = 125

2A + B = 125/0.5

2A + B = 250 (2)

On solving (1) & (2) we get,

A = 50 g/mol

B = 150 g/mol