NEET Solutions — Set 18

Question 1

A 0.6 M solution of NaCl (molar mass = 58.5 g/mol) has a density of 1.05 g/mL. What is the molality of the solution if 200 mL is prepared?

Accepted Answer

Moles = $ 0.6 \times 0.2 = 0.12 \, \text{mol} $. Mass of NaCl = $ 0.12 \times 58.5 = 7.02 \, \text{g} $. Total mass = $ 200 \times 1.05 = 210 \, \text{g} $. Mass of water = $ 210 - 7.02 = 202.98 \, \text{g} = 0.20298 \, \text{kg} $. Molality = $ \frac{0.12}{0.20298} \approx 0.591 \, \text{mol/kg} $.

Question 2

A non-volatile solute lowers the vapor pressure of water from 20 mm Hg to 19.8 mm Hg. What is the mole fraction of the solute?

Accepted Answer

Relative lowering = $ \frac{p^0 - p}{p^0} = x_{\text{solute}} $. $ \frac{20 - 19.8}{20} = \frac{0.2}{20} = 0.01 $.

Question 3

A gas has a Henry’s law constant of 300 bar at 25°C. If 0.01 mole of the gas is dissolved in 500 g of water at 6 bar, what is the molality of the solution?

Accepted Answer

Molality = $ \frac{\text{moles}}{\text{mass of solvent in kg}} = \frac{0.01}{0.5} = 0.02 \, \text{mol/kg} $. Cross-check: $ x = \frac{6}{300} = 0.02 $, consistent with molality for dilute solutions.

NEET Chemistry: Solutions — Practice Set 18

Q1. A 0.6 M solution of NaCl (molar mass = 58.5 g/mol) has a density of 1.05 g/mL. What is the molality of the solution if 200 mL is prepared?

Q2. A non-volatile solute lowers the vapor pressure of water from 20 mm Hg to 19.8 mm Hg. What is the mole fraction of the solute?

Q3. A gas has a Henry’s law constant of 300 bar at 25°C. If 0.01 mole of the gas is dissolved in 500 g of water at 6 bar, what is the molality of the solution?

Q4. A solution of a non-volatile solute in water boils at 100.52°C at 1 atm. What is the molality of the solution? (\( K_b = 0.52 \, \text{K kg mol}^{-1} \))

Q5. A gas has a Henry’s law constant of 50 bar. If its mole fraction in a solution is 0.04, what is its partial pressure above the solution?

Q6. A solution of a gas in water has a mole fraction of 0.03 at a partial pressure of 4.5 bar. If the temperature increases and the Henry’s law constant becomes 200 bar, what is the new partial pressure?

Q7. The solubility of a gas in water follows Henry’s law with a constant of 150 bar. If the gas’s partial pressure decreases from 6 bar to 3 bar, by what factor does its solubility decrease?

Q8. The osmotic pressure of a solution containing 5 g of a solute in 1 L of solution at 27°C is 2.46 atm. What is the molar mass of the solute? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

Q9. A 0.15 M solution of a solute in 500 mL of water has an osmotic pressure of 1.476 atm at 27°C. If the solute dissociates into 2 ions, what is the degree of dissociation? (\( R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \))

Q10. A solution of two volatile liquids has vapor pressures of 200 mm Hg and 300 mm Hg for pure components. If the total vapor pressure is 240 mm Hg, what is the mole fraction of the second component in the vapor phase?