10 mL of a solution of `NaCl` gave on evaporation 0.93 g of the mixed salt which gave `1.865` of `AgCl` by reacting with `AgNO_(3)` solution. Calculate the quantity of `NaCl` in 10 mL of the solution.

Explanation

The chemical equation for the reaction :
`AgNO_(3)+underset(23+35.5(=58.5g))(NaCl)rarrunderset(108+35.5)(AgCl)+NaNO_(3)`
Let the mass of `NaCl and KCl` in the mixture be respectively a g and b g.
`:. A +b = 0.93` (given)
Let us find `AgCl` formed formed on reacting `NaCl and KCl` with `AgNO_(3)` solution
58.5 of `NaCl` give `AgCl = 143.5 g`
`:. ag` of `NaCl` will give `AgCl = ((143.5 g))/((58.5g))x(ag)`
Similarly, 74.5 g of KCl give `AgCl = 143.5 g`
b g of KCl will give `AgCl=((143.5g))/((74.5g))xx(bg)`
But mass of `AgCl` actually formed = 1.865 (given)
`:.(143.5xxa)/(58.5)+(143.5xxb)/(74.5)=1.865,(143.5xxa)/(58.5)+(143.5(0.93-a))/(74.5)=1.865`
`2.453a+1.93(0.93-a)=1.865,2.453a+1.795-1.93a=1.865`
`0.523a=0.07 or a=(0.7)/(0.523)=0.14`
Mass of NaCl in the mixture = 0.14 g
Mass of KCl in the mixture `= (0.93 - 0.14) = 0.79 g`.