Question

11.2g of mixture of MCl and back gave 28.7g of white ppt with access of agno3 solution. 11.2g of same mixture on heating gave a gas that on passing into agno3 solution gave 14.35g of white ppt. Henve???

Answer 1

here's the answer and hope this helps you.

Answer 2

Answer:
The total number of moles in the mixture is 0.2.
Explanation:
Let we take MCl= x g,
NaCl = (11.2−x) g

MCl+AgNO3 ⟶ MNO3+AgCl↓
​
(M+35.5)g MCl gives 143.5g AgCl
x g MCl gives (M+35.5) / 143.5x g AgCl

NaCl +AgNO3 ⟶ NaNO3+ AgCl↓
​
(11.2−x)g NaCl gives 58.5 /143.5(11.2−x) g AgCl

(M+35.5 /143.5x ) + (58.5 / 143.5(11.2−x)) =28.7 according to the question

AgCl formed only by MCl
Hence from the above expression,
M+35.5 /143.5x = 14.35
Also ,
58.5 /143.5(11.2−x) =14.35
When we solve the above equation x=5.35
On substituting the value of x in M+35.5 /143.5x = 14.35
We get M =18
In the mixture
MCl =5.35g
NaCl =5.85g

So molar mass of MCl= 18+35.5 = 53.5g/mol
Moles of MCl= 53.5 /5.35
​ = 0.1

Moles of NaCl= 58.5 / 5.85
​ =0.1

Mole fraction of MCl= 0.5
Mole fraction of NaCl= 0.5
The total number of moles =0.2 mol