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..

Options

A. `18.25N`
B. `1.138N`
C. `0.569N`
D. `0.5N`

Answer

C

Explanation

`w = 0.5 xx 36.5 xx 1 = 18.25gr`.
Remaining wt. of `HCl = 18.25 - 2.67 = 15.58g`
`N =(w)/(GEW) xx (1000)/(V("in ml"))`
`N = (15.58)/(36.5) xx (1000)/(750) = 0.569N`