Question

5.6g of metal oxide when treated with excess of HCl gave 11.1g of metal chloride equivalent weight of metal is​

Answer 1

Let atomic mass of metal is M and valency is x.

a/c to question,

5.6 g of metal oxide when treated with excess of HCl gave 11.1g of metal chloride.

i.e., M2Ox + 2xHCl ⇔2MClx + xH2O

here it is clear that, 1 mole of metal oxide gives 2 mole of metal chloride.

so, (2M + 16x)g of metal oxides gives 2 × (M + 35.5x ) g of metal chloride.

so, 5.6g of metal oxide gives 2(M + 35.5x)/(2M + 16x) × 5.6g of metal chloride.

but given 11.1g of metal chloride.

so, 2(M + 35.5x)/(2M + 16x) × 5.6 = 11.1

⇒(2M + 71x)/(2M + 16x) = 11.1/5.6

⇒(2M + 71x)/(2M + 16x) ≈ 2

⇒2M + 71x = 4M + 32x

⇒71x - 32x = 4M - 2M = 2M

⇒39x = 2M

⇒M/x = 19.5

hence, equivalent weight = atomic mass/valency = M/x = 19.5

Answer 2

$\huge\bold\purple{Answer:-}$

so, (2M + 16x)g of metal oxides gives 2 × (M + 35.5x ) g of metal chloride.

so, 5.6g of metal oxide gives 2(M + 35.5x)/(2M + 16x) × 5.6g of metal chloride.

but given 11.1g of metal chloride.

so, 2(M + 35.5x)/(2M + 16x) × 5.6 = 11.1

⇒(2M + 71x)/(2M + 16x) = 11.1/5.6

⇒(2M + 71x)/(2M + 16x) ≈ 2

⇒2M + 71x = 4M + 32x

⇒71x - 32x = 4M - 2M = 2M

⇒39x = 2M

⇒M/x = 19.5

hence, equivalent weight = atomic mass/valency = M/x = 19.5