Question

Element A and B form two compounds B2A3 and B2A 0.05 moles of B2A3 weights 9 g and 0.10 moles of B2A weights 10 g then atomic weight of A and B

Answer 1

Answer:-

Given:

Mass of B2A3 in grams(m) = 9 g

Number of moles of B2A3 (n) = 0.05

We know that,

Number of moles (n) = m/M where M is the molar mass or atomic weight.

→ 0.05 = 9/M (B2A3)

→ M = 9/0.05

→ M (B2A3) = 180 g/mol

→ Atomic weight of B2A3 = 180

→ 2(B) + 3(A) = 180

→ 3A + 2B = 180 -- equation (1)

Similarly,

0.10 = 10/M (B2A)

→ M = 10/0.10

→ M (B2A) = 100 g/mol.

→ Atomic weight of B2A = 100

→ A + 2B = 100 -- equation (2)

Multiply equation (2) by 3 and subtract equation (1) from (2).

→ 3(A + 2B) - (3A + 2B) = 3(100) - 180

→ 3A + 6B - 3A - 2B = 120

→ 4B = 120

→ B = 120/4

→ B = 30 g/mol.

Substitute B Value in equation (1)

→ 3A + 2(30) = 180

→ 3A = 180 - 60

→ 3A = 120

→ A = 120/3

→ A = 40 g/mol

Hence, The atomic weights of A and B are 40 g/mol and 30 g/mol.