1.

How much AgClwill be formed byadding 1.70 " g of " AgNO_(3) in 200 mL of 5 N HCl solution ? (Ag = 108, N = 14 , O= 16)

Answer»

Solution :Equivalent of `AgNO_(3) = (1.70)/(170) = 0.71 `
m.e of HCl solution = `5 xx 200 = 1000`
` :. ` equivalent of HCl solution ` = 1000/(1000) = 1 `
Since equivalent of `AgO_(3)`is less than the eq. of HCl ,
equivalent of AgCl = eq. of `AgNO_(3) = 0.01 `
` :. ` wt. of AgCl = `0.01 xx 1435`
` = 1.435 G `


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