Question

2.66 g of a mixture of KCl and NaCl gave, 5.74 g of dry silver chloride, on treatment with silver nitrate solution. The molar ratio of NaCl to KCl is, (Molar mass of KCl is 74.5 gmol-1, NaCl is 58.5 gmol-1 and AgCl is 143.5 gmol-1)​

Answer 1

Answer:

The answer to this question is 1.021 : 1.639

Explanation:

Following is the reaction of silver nitrate and sodium chloride:

NaCl + $AgNO_{3}$ → $NaNO_{3}$ + AgCl

1 mol      1 mol    

58.5gmol-1    143.5gmol-1

We can see that 1 mol sodium chloride is reacting with 1 mol of silver nitrate.

Therefore,

If x is the amount of NaCl produced, then it will be as follows:

x gram NaCl = 143.5g / 58.5g × x g $AgNO_{3}$

Similarly for KCl and silver nitrate we have the following reaction:

KCl     + $AgNO_{3}$ → $KNO_{3}$  + AgCl

1 mol      1 mol

74.5gmol-1  143.5gmol-1

Therefore total 2.66 g mixture of the KCl and the NaCl was present hence,

if xg is NaCl then KCl = 2.66g - xg

2.66 -x g KCl = 143.5g/ 74.5g × (2.66 - x)g $AgNO_{3}$

Total 5.74g of dry silver chloride was produced therefore

2.45 × x + 1.92 (2.66 - x) = 5.74

∴ x = 1.021gmol-1

2.66 - x = 2.66 - 1.021 = 1.639 gmol-1

Molar ratio = NaCl : KCl = 1.021 : 1.639

Therefore this is the required ratio.

Answer 2

Given:

• Molar Mass of KCl = 74.5 $gmol^{-1}$

• Molar Mass of NaCl = 58.5 $gmol^{-1}$

• Molar Mass of AgCl = 143.5 $gmol^{-1}$

To Find:

• The Molar ratio of NaCl to KCl.

Solution:

• Let us first give a  reaction between silver nitrate and KCl.
• $KCl + AgNO_3$ →$KNO_3 + AgCl$
• In the above reaction, Silver Nitrate is used because the two compounds mentioned are treated with the silver nitrate solution.
• We already have the molar masses of KCl and NaCl
• 2.66 g of  both the compounds were present,
• Hence we can say that, let Xg = Nacl
• So we get KCl = 2.66g - Xg
• 2.66g - Xg KCl = 143.5/74.5 * (2.66 - Xg)$AgNO_3$
• We get, 1.92*(2.66 - Xg)$AgNO_3$
• By the given data, 5.74 g of dry silver chloride was present.
• We can write, 2.45*X + 1.92*(2.66-X) = 5.74
• 2.45X + 5.10 - 1.92X = 5.74
• 2.45X - 1.92X = 5.74 - 5.10
• 0.53X = 0.64
• X = 1.20
• Substitue X in the equation, 2.66g - Xg
• 2.66 - 1.20 = 1.46g

The Molar Ratio of NaCl and KCl is 1.20 : 1.46