Question

$11.2$ g of mixture of $MCl$ (volatile) and $NaCl$ gave $28.7$ g of white precipitate with excess of $AgN{O}_3$ solution. $11.2$ g of same mixture on heating gave a gas that on passing into $AgN{O}_3$ solution gave $14.35$ g of white precipitate. $MCl$ and $NaCl$ have the molar ratio $x:y$.
The value of $10(x+y)$ is :

Answer 1

$LetMCl=xg,NaCl=(11.2-x)g$
$MCl+AgN{O}_3⟶MN{O}_3+\underset{whiteppt}{AgCl\downarrow }$
$(M+35.5)gMCl$ gives $=143.5gAgCl$
$xgMCl$ gives $=\frac{143.5x}{(M+35.5)}gAgCl$ .....(i)
$\underset{\left(58.5g\right)}{NaCl}+AgN{O}_3⟶NaN{O}_3+\underset{143.5g}{AgCl\downarrow }$
$(11.2-x)gNaCl$ gives $=\frac{143.5(11.2-x)}{58.5}gAgCl$ ... (ii)
$\frac{143.5x}{M+35.5}+\frac{143.5(11.2-x)}{58.5}=28.7$ (given) ...( iii)
On heating MCl vaporises (being volatile).
Thus, AgCl is due to MCl only.
Thus, $\frac{143.5x}{M+35.5}=14.35$ ... (iv)
Thus, from (iii),
$\frac{143.5(11.2-x)}{58.5}=14.35$
Thus, gives $x=5.35$
From (iv), $M=18$
$MCl$ in mixture $=5.35g$
$NaCl$ in mixture $=5.85g$
Thus, ionic mass of $M^{+}$ is $18$.
Also, molar mass of $MCl=53.5gmo{l}^{-1}$
Moles of $MCl=\frac{5.35}{53.5}=0.1$
Moles of $NaCl=\frac{5.85}{58.5}=0.1$
Total moles $=0.2$
Mole fraction of $MCl=0.5$
Mole fraction of $NaCl=0.5$

Hence, the toatl number of moles $=0.2$ mol

Therefore, the value of $10(x+y)=10\times 0.2=2$ mol