Question

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1.331 g of a mixture of KCl and NaCl gave, on treatment with silver nitrate ($\sf AgNO_3$) solution, 2.876 g of dry silver chloride.
Find the percentage composition of mixture?

Answer given :- KCl = 55.47%

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Answer 1

hey there

refer to attachment

Answer 2

Answer:

➟ KCl + AgNO3--> AgCl + KNO3 (lets look for molar weighs)

➟ KCl = 39.1+35.5=74.6g and AgCl 107.9+35.5=143.4g (So ratio of molar masses AgCl/KCl=1.925

➟ NaCL+AgNO3---> AgCl + NaNO3

Molar mass NaCl=23+35.5=58.5g and ratio molar masses AgCl/NaCl=143.4/58.5=2.451

suppose we have x is proportion NaCl ,the proportion of KCl is 1-x

the ratio 2.876/1.331=2.161

So,

➟2.161= 2.451x + 1.925(1-x)

➟2.161-1.925= 2.451x-1.925x

➟0.526x=0.236

x=0.449 or 44.9% of NaCl and 100-44.9= 55.1% KCl