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10 mL of an HCl solution gave 0.1435 g o...

10 mL of an HCl solution gave `0.1435` g of AgCl when treated with excess of `AgNO_(3)` .The normality of the resulting solution is

Text Solution

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`"Cl present in 0.1435 g AgCl"=(35.5)/(143.5)xx0.1435g=0.0355g`
`"HCl containing 0.0355 g Cl"=(36.5)/(35.5)xx0.0355g=0.0365g=(0.0365)/(36.5)"g eq."="0.001 g eq."`
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Knowledge Check

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