10 mL of HCl solution gave 0.1435 g of AgCl when treated with excessof – InterviewSolution

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1.	10 mL of HCl solution gave 0.1435 g of AgCl when treated with excessof AgNO_(3). Find the normality of the acid solution [Ag = 108]
Answer» SOLUTION :`"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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