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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) |
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Answer» `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)` |
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