An aqueous sodium hydroxide solution contains, 80ppm NaOH. If one ml of the solution is mixed with 99 ml water, the molarity of the resultant solution is `x xx10^(-5)` what is x? (assume d=gm/cc)

Answer

2

Explanation

ppm = `(w_(1))/(w_(2))xx10^(6)=80`
`M=(w_(1))/(40xx((w_(2))/(1000)))(because w_(2)=V_("soln (mL)"))`
`=(w_(1))/(w_(2))xx(1000)/(40)=(80xx10^(-6)xx1000)/(40)=2xx10^(-3)M`
`M_(1)V_(1)=M_(2)V_(2)`
`1xx2xx10^(-3)=100xxM_(2)impliesM_(2)=2xx10^(-5)`