Question

1 gram of a mixture of potassium chloride and sodium chloride on treatment with excess of silver nitrate have 2 gram of silver chloride what is the composition of original mixture

Answer 1

Set up this equation:

(x) (grams Cl from NaCl) + (1-x) (grams Cl from KCl) = total grams chloride

x = grams of NaCl in original mixture

1-x = grams of KCl in original mixture

grams Cl from NaCl = 35.5/58.4

grams Cl from KCl = 35.5/74.6

total grams chloride in AgCl = (2 g) (35.5/143.3)

The numbers in the denominators are the molar masses of NaCl, KCl and AgCl.

The three ratios are called "gravimetric factors."

(x) (35.5/58.4) + (1-x) (35.5/74.6) = (2 g) (35.5/143.3)

Solve:

(x) (35.5/58.4) + (1-x) (35.5/74.6) = (2g) (35.5/143.3)

0.6079x + 0.4759 - 0.4759x = 0.4955

0.132x = 0.0196

x = 0.148 g

This is the mass of NaCl in the original mixture.

This computes to 15% of the original mixture.

Answer 2

Answer:

Explanation:

Set up this equation:

(x) (grams Cl from NaCl) + (1-x) (grams Cl from KCl) = total grams chloride

x = grams of NaCl in original mixture

1-x = grams of KCl in original mixture

grams Cl from NaCl = 35.5/58.4

grams Cl from KCl = 35.5/74.6

total grams chloride in AgCl = (2 g) (35.5/143.3)

The numbers in the denominators are the molar masses of NaCl, KCl and AgCl.

The three ratios are called "gravimetric factors."

(x) (35.5/58.4) + (1-x) (35.5/74.6) = (2 g) (35.5/143.3)

Solve:

(x) (35.5/58.4) + (1-x) (35.5/74.6) = (2g) (35.5/143.3)

0.6079x + 0.4759 - 0.4759x = 0.4955

0.132x = 0.0196

x = 0.148 g

This is the mass of NaCl in the original mixture.

This computes to 15% of the original mixture.