Question

A solution is prepared by mixed by mixing 0.10L of 0.14M sodium Chloride with 0.21L of a 0.19M MgCl2 solution. What volume of 0.21M silver Nitrate is required to precipitate all the Cl- ion in the solution AgCl?​

Answer 1

Answer:0.25$L$

Explanation:

If $M$ is the molarity and $V$ is the molarity of a given solution,

number of moles of the solute in the solution is $MV$.

Number of moles of $Cl^{-1}$ ions in the initial solution is the sum of number of moles of $Cl^{-1}$ ions  in $NaCl$ and $MgCl_{2}$

Number of moles of $Cl^{-1}$ ions in $NaCl$ is $0.14\times 0.1=0.014moles$

number of moles of $Cl^{-1}$ ions  in $MgCl_{2}$ is $0.19\times 0.2=0.038moles$

Number of moles of $Cl^{-1}$ ions in the initial solution is $0.014+0.038=0.052moles$

So,$0.052moles$ of $Ag^{+}$ ions are required to displace $0.052moles$ of $Cl^{-}$ ions.

So,number of moles of $AgCl$ required is $0.052moles$

$V=\frac{n}{M}$

So,$V=\frac{0.052}{0.21}=0.25L$