How many moles of `BaCl_(2)` would be needed to make 250 ml of a solution having the same concentration of `Cl^(-)` as one containing 1.17 g NaCl per 100ml?

Options

A. 0.1
B. 0.01
C. 0.05
D. 0.025

Answer

D

Explanation

`[Cl^(-)]_(NaCl)=(1.17)/(58.5)xx(1000)/(100)=[Cl^(-)]_(BaCl_(2))`
`therefore [Cl^(-)]_(BaCl_(2))=("mole "Cl^(-))/((250)/(1000))=0.2`
Mole of `Cl^(-)=0.2 xx(1)/(4)`
`therefore` mole of `BaCl_(2) = 0.2 xx(1)/(4)xx(1)/(2)=0.025`