Upon heating 1 L of 2 N HCl solution, 36.5 g of HCl is lost and the volume of solution resuces to 800 mL. Calculate . (a). The normality of the resultant solution.
(b). The number of equivalents of HCl in 100 " mL of " the original solution..

Explanation

m" Eq of "`HCl=NxxV(mL)=2xx1000=2000mEq`
`(W)/(Ew)xx10^(3)=2000`
`(Wxx1000)/(36.5)=2000`
`W_(HCl)=2xx36.5`
(a). Since. 36.5 g of HCl is lost on heating and volume becomes 800 mL
`W_(HCL)` left `=2xx36.5-36.5=36.5g`
m" Eq of "HCl left `=(Wxx10^(3))/(Ew)=(36.5xx1000)/(36.5)`
`=1000mEq`
Normality of resultant solution
`=(mEq)/(V(mL))=(1000)/(800)=1.25N`
(b). m" Eq of "HCl in 1 L of original solution `=2000`
m" Eq of "HCl in 100 " mL of " original solution is 200 mEq
`=200xx10^(-3)=0.2Eq`.