Upon heating a litre of a $\frac{N}{2}$ $H C l$ solution, $2.675$ g of hydrogen chloride is lost and the volume of the solution shrinks to $750$ mL. Calculate (i) the normality of the resultant solution (ii) the number of milliequivalents of $H C l$ in $100$ mL of the original solution.

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Solution

Normality of $H C l = \frac{1}{2} N$
Amount of solution = 1 litre
Mass of $H C l = \frac{1}{2} N \times 1 l i t r e \times 36.5 = 18.25$
After heating mass of $H C l = 18.250 - 2.675 = 15.575$
After heating volume = $750 m l$
(i) Normality = $\frac{15.575 / 36.5}{0.75} = 0.569 N$
(ii) Number of millequivalent = $\frac{N}{2} \times 100 m l$
$= 50$ mill eq

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Upon heating a litre of a N2 HCl solution, 2.675g of hydrogen chloride is lost and the volume of the solution shrinks to 750mL. Calculate (i) the normality of the resultant solution (ii) the number of milliequivalents of HCl in 100mL of the original solution.

Normality of HCl=12NAmount of solution = 1 litreMass of HCl=12N×1litre×36.5=18.25After heating mass of HCl=18.250−2.675=15.575After heating volume = 750ml(i) Normality = 15.575/36.50.75=0.569N(ii) Number of millequivalent = N2×100ml=50 mill eq

Upon heating a litre of a $\frac{N}{2}$ $H C l$ solution, $2.675$ g of hydrogen chloride is lost and the volume of the solution shrinks to $750$ mL. Calculate (i) the normality of the resultant solution (ii) the number of milliequivalents of $H C l$ in $100$ mL of the original solution.