Question

In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO₃, which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine)

(a) X₂Cl₂

(b) XCl₂

(c) XCl₄ 

(d) X₂Cl

Answer 1

Ans. (b) XCl₂

Sol. molarity = moles\volume(L)

0.05 = moles/10 x 1000

therefore, moles of chloride = 0.05 x 10/1000 = 0.5/1000 = 5 x 10^-5

so gram equivalents of chloride = moles x ’n’ factor = n x  5 x 10^-5

here n factor means XCln i.e it is the valence factor.

similarly,

moles of AgNO₃ = molarity x volume = 0.1 x 10/1000 = 1/1000 = 10^-4

gram equivalents of AgNO₃ = moles x ’n’ factor = 1 x 1/1000 =1/1000 =  10^-4

gram equivalents of chloride=gram equivalents of AgNO₃ 

 n x  5 x 10^-5 = 10^-4

0.5n = 1

n = 2

therefore, formula of compound = XCl₂