Exams › JEE Main › Chemistry

A solution is made by mixing 36 g of water with 46 g of glycerine, C3H5(OH)3. Find the mole fraction of glycerine. (Molar masses: water = 18, glycerine = 92)

1. 0.20
2. 0.46
3. 0.36
4. 0.40

Correct answer: 0.20

Solution

With 2 mol water and 0.5 mol glycerine, x(glycerine) = 0.5/(2 + 0.5) = 0.5/2.5 = 0.20.

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 →