1.

If 0.20 g chloride of a certain metal, when dissolved in water and treated with excess of AgNO_(3) yields 0.50 g of AgCl, the equivalent mass of the metal is (Ag=108, Cl = 35.5)

Answer»

21.9
20.04
40.08
43.80

Solution :`(E+"EQ. mass of" CL^(-))/("Eq. mass of AG + Eq. mass of" Cl^(-))=(E+35.5)/(108+35.5)`
`=(0.2)/(0.5) implies E = 21.90`


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