JEE Main 2022 · 26 Jul · Shift 1 · Chemistry Q85

← JEE Main 2022 · 26 Jul · Shift 1

ChemistryInteger+4 / -1Question 85 of 90

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 $x\times10^{-2}$ M. (Nearest Integer)

(Molar mass of nitric acid is 63 g mol $^{-1}$ )

Your answer

Answer and worked solution

Official answer: 54

Moles $0.4-\frac{11.5}{63}=0.2175$ in 0.4 L: 0.54 M.

Answer: 54

Topics:Concentration terms

← Previous question Next question →

JEE Main 2022 · 26 Jul · Shift 1, Chemistry question 85