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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

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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

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