JEE Chemistry States Of Matter MCQs Set 09

Answer: (b) Surface tension

Explanation:
1. A liquid tends to minimise its surface area because of surface tension.
2. For a given volume a sphere has the smallest surface area, so free drops become spherical.

Teacher's Note:
1. Surface tension arises from unbalanced forces on surface molecules.
2. It also explains capillary rise and the shape of soap bubbles.

Answer: (b) the polarisability of the interacting particles

Explanation:
1. London (dispersion) forces arise from instantaneous induced dipoles.
2. The more easily the electron cloud is distorted (greater polarisability), the stronger the force.

Teacher's Note:
1. Dispersion forces act between all molecules, including non-polar ones.
2. Larger molecules are more polarisable and have stronger London forces.

Answer: (b) Pressure increases

Explanation:
1. At constant volume, p/T is constant (Gay-Lussac's law).
2. A rise in temperature raises the average kinetic energy, so molecules hit the walls harder and more often, increasing the pressure.

Teacher's Note:
1. Pressure is directly proportional to absolute temperature at constant V.
2. Never use Celsius temperature in gas law ratios.

Answer: (a) O2, N2, H2, He

Explanation:
1. A gas can be liquefied by pressure only when it is below its critical temperature.
2. The gas with the highest critical temperature liquefies most easily.
3. The order of Tc is O2 > N2 > H2 > He, which is the order of liquefaction.

Teacher's Note:
1. Higher critical temperature means stronger intermolecular forces.
2. Helium is the most difficult gas to liquefy.

Answer: (a) It decreases

Explanation:
1. At a higher temperature the molecules have more kinetic energy.
2. This helps overcome the intermolecular forces, so the liquid flows more easily and viscosity decreases.

Teacher's Note:
1. The viscosity of gases, in contrast, increases with temperature.
2. Viscosity of liquids falls roughly exponentially with temperature.

Answer: (a) It decreases

Explanation:
1. A higher temperature raises the kinetic energy of the molecules and weakens the net inward pull on surface molecules.
2. So surface tension decreases and becomes zero at the critical temperature.

Teacher's Note:
1. Surface tension falls as temperature rises.
2. Detergents also lower surface tension.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. Strong intermolecular forces dominate in solids and liquids, thermal energy dominates in gases.
2. The reason states exactly how the balance operates, so it explains the assertion.

Teacher's Note:
1. Increasing temperature moves matter from solid to liquid to gas.
2. Increasing pressure favours the condensed states.

Answer: (c) A is true but R is false

Explanation:
1. The assertion is the correct definition of boiling point.
2. At high altitude the atmospheric pressure is low, so the reason is false.
3. That is why water boils below 100 degrees C on mountains.

Teacher's Note:
1. Normal boiling point is measured at 1 atm.
2. Pressure cookers raise the boiling point by raising the pressure.

Answer: (d) A is false but R is true

Explanation:
1. Liquids tend to have the minimum number of molecules at the surface, to minimise the surface energy, so A is false.
2. Small drops are spherical because a sphere has minimum surface area, so R is true.

Teacher's Note:
1. Minimum surface area means minimum surface energy.
2. Surface tension is the cause of both observations.

Answer: (c) 1 : 1.128 : 1.225

Explanation:
1. \(u_{mp} = \sqrt{2RT/M}\), \(u_{av} = \sqrt{8RT/\pi M}\), \(u_{rms} = \sqrt{3RT/M}\).
2. Taking ratios: \(\sqrt{2} : \sqrt{8/\pi} : \sqrt{3} = 1.414 : 1.596 : 1.732 = 1 : 1.128 : 1.225\).

Teacher's Note:
1. Memorise the order u_mp < u_av < u_rms.
2. All three are proportional to the square root of T/M.

Answer: (d) a for \(\ce{Cl2}\) is larger but b is smaller than for \(\ce{C2H6}\)

Explanation:
1. Ease of liquefaction depends on the van der Waals constant a (attractive forces), and a larger a means easier liquefaction.
2. Cl2 has a = 6.49 and b = 0.056, while ethane has a = 5.56 and b = 0.064.
3. So a for Cl2 is larger and b is smaller.

Teacher's Note:
1. Constant a measures intermolecular attraction; b is the effective volume of molecules.
2. Gases with higher a have higher critical temperature.

Answer: (b) directly proportional to the square root of temperature

Explanation:
1. \(u_{rms} = \sqrt{3RT/M}\).
2. So the velocity is proportional to the square root of the absolute temperature.

Teacher's Note:
1. Doubling T increases the speed by a factor of 1.41.
2. At the same T, lighter gases move faster.

Answer: (a) \(1.42\times10^{-3}\)

Explanation:
1. Total water = 0.1 g = 0.1/18 = \(5.56\times10^{-3}\) mol.
2. Moles that can exist as vapour: n = pV/RT = (26.7/760)(1.0)/(0.0821 x 300) = \(1.43\times10^{-3}\) mol.
3. This is less than the total water, so equilibrium is reached with \(1.42\times10^{-3}\) mol (approx.) in the vapour phase.

Teacher's Note:
1. Check whether all the liquid can evaporate before using the vapour pressure.
2. Equilibrium vapour pressure fixes the moles in the vapour.

Answer: (c) The fraction of molecules with the most probable speed increases

Explanation:
1. The area under the curve is the total fraction of molecules (one) and remains unchanged.
2. The curve flattens and broadens and the peak shifts to higher speed.
3. The peak height, i.e. the fraction of molecules at the most probable speed, decreases, so (c) is not true.

Teacher's Note:
1. At higher T the curve shifts right and flattens.
2. The area under the curve stays constant.

Answer: (b) 313/293

Explanation:
1. Average kinetic energy is proportional to absolute temperature.
2. Ratio = 313 K / 293 K.

Teacher's Note:
1. Convert to kelvin before taking ratios.
2. The kinetic energy rises only by about 7 per cent.

Answer: (d) the volume occupied by the molecules

Explanation:
1. Constant b is the excluded volume, which is four times the actual volume occupied by one mole of the gas molecules.
2. Constant a measures the attractive forces.

Teacher's Note:
1. Large molecules have large b.
2. Units of b are L/mol.

Answer: (c) in a straight line

Explanation:
1. Ideal gas molecules exert no forces on one another except in collisions.
2. Between collisions they move in straight lines at constant speed.

Teacher's Note:
1. Kinetic theory assumes no intermolecular forces.
2. Collisions are perfectly elastic.

Answer: (a) 1.67

Explanation:
1. For a monatomic ideal gas, Cv = 3R/2 and Cp = 5R/2.
2. The mixture of two monatomic gases is also monatomic, so Cp/Cv = 5/3 = 1.67.

Teacher's Note:
1. Gamma = 5/3 for monatomic and 7/5 for diatomic gases.
2. Mixing monatomic gases does not change gamma.

Answer: (c) The mass of a gas cannot be determined by weighing the container in which it is enclosed

Explanation:
1. Gases have mass and so the mass of a gas can be found from the weight difference between a filled and an empty container.
2. The other three statements are correct.

Teacher's Note:
1. Density of a gas is found this way.
2. Gases occupy the full volume of the container.

Answer: (d) 0 degrees C

Explanation:
1. Charles' law: \(V_t = V_0(1 + t/273.15)\).
2. The increase per degree is 1/273.15 of the volume at 0 degrees C.

Teacher's Note:
1. This is the original statement of Charles' law.
2. Volume is zero at absolute zero in the ideal limit.

Answer: (b) 4

Explanation:
1. Moles of O2 = 3.2/32 = 0.1; moles of H2 = 0.2/2 = 0.1.
2. Total moles = 0.2.
3. p = nRT/V = 0.2 x 0.0821 x 273/1.12 = 4.0 atm.

Teacher's Note:
1. At STP one mole occupies 22.4 L.
2. Here 0.2 mol x 22.4 L / 1.12 L = 4 atm.

Answer: (b) 4

Explanation:
1. Pressure due to B = 3 - 2 = 1 bar.
2. In the same V and T, moles are proportional to pressure: \(n_B/n_A = 1/2\).
3. \(n = w/M\), so \(M_B/M_A = (w_B/w_A)(n_A/n_B) = 2 \times 2 = 4\).

Teacher's Note:
1. Partial pressure is proportional to moles at fixed V and T.
2. Dalton's law of partial pressures applies.

Answer: (d) 44

Explanation:
1. Graham's law: \(r_1/r_2 = \sqrt{M_2/M_1}\).
2. 127/100 = \(\sqrt{71/M}\), so M = 71/(1.27)^2 = 44.

Teacher's Note:
1. Lighter gases diffuse faster.
2. Equal time means rate is proportional to volume.

Answer: (a) 56

Explanation:
1. For \(\ce{C_xH_y}\), 4 volumes of CO2 means x = 4.
2. Oxygen needed = x + y/4 = 6, so y = 8.
3. The hydrocarbon is C4H8 with molar mass 56.

Teacher's Note:
1. Use Gay-Lussac's law of combining volumes.
2. Oxygen requirement is x + y/4 volumes.

Answer: (d) 22

Explanation:
1. Moles of CO2 = 44/44 = 1; moles of H2 = 44/2 = 22.
2. At constant V and T, pressure is proportional to moles, so p(H2) = 22 atm.

Teacher's Note:
1. Equal mass of a lighter gas means many more moles.
2. Use p proportional to n.

Answer: (b) 16

Explanation:
1. Rate is inversely proportional to the square root of M.
2. \(2 = \sqrt{64/M}\), so M = 64/4 = 16.

Teacher's Note:
1. Four times lighter gas diffuses twice as fast.
2. Graham's law of diffusion.

Answer: (a) 327

Explanation:
1. Average speed is proportional to \(\sqrt{T/M}\).
2. \(T_1/64 = 300/32\), so \(T_1 = 600\) K.
3. In Celsius, 600 - 273 = 327 degrees C.

Teacher's Note:
1. Heavier gases need a higher temperature for the same speed.
2. Convert the final answer back to Celsius.

Answer: (c) 500

Explanation:
1. In an open vessel p and V are constant, so nT is constant.
2. Three-fifths of the air remains, so \(n_1 T_1 = (3/5)n_1 T_2\).
3. \(T_2 = 300 \times 5/3 = 500\) K.

Teacher's Note:
1. In an open vessel moles escape as temperature rises.
2. Use nT = constant.

Answer: (c) 32

Explanation:
1. Rate of Y is twice the rate of X.
2. \(r_Y/r_X = \sqrt{M_X/M_Y} = 2\), so \(M_Y = 128/4 = 32\).

Teacher's Note:
1. Faster gas is lighter.
2. Rates are volumes per unit time.