When 800 mL of 0.5 M nitric acid is heated | vedclass

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(D) Initial moles of $HNO_3$ = $Molarity 	imes Volume(L) = 0.5 	imes 0.8 = 0.4 \ mol$.Moles of $HNO_3$ evaporated = $\frac{Mass}{Molar \ mass} = \fr

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(D) Initial moles of $HNO_3$ = $Molarity 	imes Volume(L) = 0.5 	imes 0.8 = 0.4 \ mol$.Moles of $HNO_3$ evaporated = $\frac{Mass}{Molar \ mass} = \frac{11.5 \ g}{63 \ g \ mol^{-1}} \approx 0.1825 \ mol$.Moles of $HNO_3$ remaining = $0.4 - 0.1825 = 0.2175 \ mol$.New volume of solution = $\frac{800 \ mL}{2} = 400 \ mL = 0.4 \ L$.New molarity = $\frac{Moles \ remaining}{Volume(L)} = \frac{0.2175}{0.4} = 0.54375 \ M$.Expressing in $x 	imes 10^{-2} \ M$,we get $54.375 	imes 10^{-2} \ M$.Rounding to the nearest integer,$x = 54$.

When $800 \ mL$ of $0.5 \ M$ nitric acid is heated in a beaker,its volume is reduced to half and $11.5 \ g$ of nitric acid is evaporated. The molarity of the remaining nitric acid solution is $x \times 10^{-2} \ M$. (Nearest Integer) (Molar mass of nitric acid is $63 \ g \ mol^{-1}$)

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