Exams › JEE Main › Chemistry

When 800 mL of 0.5 M nitric acid is boiled in a beaker, its volume falls to half and 11.5 g of HNO3 is lost by evaporation. The molarity of the remaining acid is x * 10⁻² M. Find the nearest integer x. (Molar mass of HNO3 = 63 g/mol.)

1. 55
2. 63
3. 50
4. 45

Correct answer: 55

Solution

Remaining moles = 0.4 - 11.5/63 = 0.2175 mol in 0.4 L, giving about 0.544 M = 55 * 10⁻² M.

Related JEE Main Chemistry questions

• A sulphuric acid solution has a concentration of 3.60 M and contains 29% H2SO4 by mass. If the molar mass of H2SO4 is 98 g mol⁻¹, what is the density of the solution in g mL⁻¹?
• Which one of the following substances is absent in clear hard water?
• Fractional distillation is chosen in a situation where
• Car radiators use an aqueous ethylene glycol solution as antifreeze. How much ethylene glycol should be mixed with 5 kg of water so that the freezing point is lowered to -0.3 °C? (K_f = 1.86 K kg mol⁻¹)
• A solution contains a non-volatile solute of molecular mass M₂. Which expression can be used to determine the molecular mass of the solute using osmotic pressure?
• At 298 K, the Henry’s law constant for oxygen is 1.4 × 10⁻³ mol L⁻¹ atm⁻¹. If oxygen is at a partial pressure of 0.5 atm, what mass of oxygen will dissolve in 100 mL of water at 298 K?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →