Question

One gram of a mixture of potassium and sodium chloride on treatment with excess of silver nitrate give 2 gram of silver chloride. What was the composition of the two salts in the original mixture?

Answer 1

Answer:

XKCI + yNaCl + (x+y)AgNO3 = xKNO3 + yNaNO3 + (x+y)AgCl

Let x moles of KCI and y moles of NaCl

present in the mixture

The moles of AgCl formed = x+y= 2/143.5

moles

Therefore

x + y = 2/143.5

(1)

Given that,

74.5x +58.5y = 1 -(2)

solving the two equations for x,y multiply first equation by 74.5 74.5x + 74.5y = 2 x 74.5/143.5 = 149/143.5 =

1.038

74.5x + 58.5y = 1

(74.5-58.5)y = 1.038 -1

y = 0.038/16 = 0.00237 moles

mass of NaCl = 0.0237 x 58.5 = 0.139 g 14% =

mass of KCI = 1-0.139 0.861 g = 86%

The composition of original mixture is 14 %

NaCl and 86% KCI

Explanation:

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