A mixture contains `NaCl` and unknown chloride `MCl`.
(a) `1 g` of this is dissolved in water, excess of acidified `AgNO_(3)` solution is added to it, so that `2.567 g` of white `ppt`. Is obtained.
(b) `1 g` of original mixture is heated to `300^(circ)C`. Some vapours come out which are absorbed in `AgNO_(3)` ( acidified) solution. `1.341 g` of white precipitate is obtained.
Find the mol.wt. of unkonwn chloride.

Explanation

Suppose the molecular weight of MCI is M .
Given that :
`{:(NaCl = MCI overset(AgNO_(3))to AgCl),("(1 - x) g x g (say) 2 . 567 g "):}`
Applying POAC for Cl atoms,
moles of Cl in NaCl + moles of Cl in MCI = moles of Cl in AgCl
` 1 xx ` moles of NaCl `+ 1 xx ` moles of MCl = `1 xx ` moles of AgCl
`(1 - x)/( 58 . 5) + (x)/( M) = (2 . 567)/( 143 . 5 )` . . . (i)
Now , futher of `300^(@) C ` MCI is supposed to undergo sublimation while NaCl does not.
We have,
`{:(MCI underset(300^(@)C)overset(AgNO_(3))to AgCl),("x g 1.341 g "):}`
Applying POAC for Cl atoms,
moles of Cl in MCI = moles of Cl in AgCl
`1 xx ` moles of MCI `= 1 xx ` moles of AgCl
`(x)/( M) = (1 . 341)/( 143 . 5) " ". . . (ii)`
From equation (i) and (ii) , we get
M = 53.5
`:.` mol. wt. of MCI = 53.5