JEE Chemistry States Of Matter MCQs Set 10

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Multiple Choice Questions

Question 1: A gas occupies 100.0 mL at 50°C and 1 atm. It is cooled at constant pressure until its volume is 50.0 mL. The final temperature is:
(a) \(161.5^\circ\)C
(b) \(-161.5^\circ\)C
(c) \(-111.5^\circ\)C
(d) \(25^\circ\)C
Show Answer & Explanation

Answer: (c) \(-111.5^\circ\)C

Explanation:
1. At constant pressure \(V_1/T_1 = V_2/T_2\), with \(T_1 = 323\) K.
2. \(T_2 = 323 \times 50/100 = 161.5\) K.
3. In Celsius, \(161.5 - 273 = -111.5^\circ\)C.

Teacher's Note:
1. Always convert to kelvin before using Charles' law.
2. The answer 161.5 is the kelvin value; the question asks for the final temperature, which is negative on the Celsius scale.

Question 2: A sample of gas at \(27^\circ\)C and 1.00 atm occupies 2.50 L. The temperature at which its pressure becomes 1.50 atm after transfer to a 2.00 L container is:
(a) 360 K
(b) 450 K
(c) 300 K
(d) 225 K
Show Answer & Explanation

Answer: (a) 360 K

Explanation:
1. \(P_1V_1/T_1 = P_2V_2/T_2\).
2. \(T_2 = 300 \times (1.5 \times 2.00)/(1.00 \times 2.50) = 360\) K.

Teacher's Note:
1. The combined gas law applies when P, V and T all change.
2. Check that the pressure rises and the volume falls, so T must rise.

Question 3: The mass of hydrogen needed to inflate a balloon to volume V at \(127^\circ\)C is (8 g of helium inflates it to 0.5 V at \(27^\circ\)C, same pressure):
(a) 6 g
(b) 3 g
(c) 1.5 g
(d) 12 g
Show Answer & Explanation

Answer: (a) 6 g

Explanation:
1. 8 g He = 2 mol, and \(n \propto V/T\) at constant pressure.
2. \(n_{H_2} = 2 \times (V/0.5V) \times (300/400) = 3\) mol.
3. Mass = \(3 \times 2 = 6\) g.

Teacher's Note:
1. At constant P, n is proportional to V/T.
2. Do not confuse the moles of He with the mass of H\(_2\).

Question 4: A mixture of gases at 760 torr contains 55.0% N\(_2\), 25.0% O\(_2\) and 20.0% CO\(_2\) by mole. The partial pressures (in torr) of N\(_2\), O\(_2\) and CO\(_2\) are respectively:
(a) 380, 190, 190
(b) 152, 190, 418
(c) 418, 190, 152
(d) 190, 418, 152
Show Answer & Explanation

Answer: (c) 418, 190, 152

Explanation:
1. Partial pressure = mole fraction \(\times\) total pressure.
2. \(0.55 \times 760 = 418\), \(0.25 \times 760 = 190\), \(0.20 \times 760 = 152\) torr.

Teacher's Note:
1. Partial pressures add up to the total: 418 + 190 + 152 = 760.
2. Percentage by mole is the same as mole fraction times 100.

Question 5: The pressure exerted by a mixture of 3.2 g of methane and 4.4 g of CO\(_2\) in a 9 dm\(^3\) flask at \(27^\circ\)C is (R = 0.0821 L atm K\(^{-1}\) mol\(^{-1}\)):
(a) 1.64 atm
(b) 0.821 atm
(c) 0.41 atm
(d) 0.082 atm
Show Answer & Explanation

Answer: (b) 0.821 atm

Explanation:
1. \(n_{CH_4} = 3.2/16 = 0.2\), \(n_{CO_2} = 4.4/44 = 0.1\), total 0.3 mol.
2. \(P = nRT/V = 0.3 \times 0.0821 \times 300/9 = 0.821\) atm.

Teacher's Note:
1. 1 dm\(^3\) is 1 L.
2. Dalton's law lets you add the moles of non-reacting gases.

Question 6: At the top of a mountain the temperature is \(-23^\circ\)C and the pressure 700 mm Hg. At the bottom they are \(27^\circ\)C and 750 mm Hg. The ratio of air density at the top to that at the bottom is:
(a) 1.07
(b) 0.93
(c) 0.89
(d) 1.12
Show Answer & Explanation

Answer: (d) 1.12

Explanation:
1. \(d = PM/RT\), so \(d \propto P/T\).
2. \(d_{top}/d_{bottom} = (700/250)/(750/300) = 2.8/2.5 = 1.12\).

Teacher's Note:
1. At the same pressure, a colder gas is denser.
2. Use kelvin: 250 K and 300 K.

Question 7: A 22.4 L container at 1 atm and \(0^\circ\)C holds Ar, O\(_2\) and SO\(_2\) such that \(P_{SO_2} = P_{O_2} + P_{Ar}\) and \(P_{O_2} = 2P_{Ar}\). The density of the mixture is nearly:
(a) 1.78 g/L
(b) 3.20 g/L
(c) 2.86 g/L
(d) 2.20 g/L
Show Answer & Explanation

Answer: (d) 2.20 g/L

Explanation:
1. Let \(P_{Ar} = p\): then \(P_{O_2} = 2p\), \(P_{SO_2} = 3p\), and \(6p = 1\) atm.
2. Mole fractions: Ar 1/6, O\(_2\) 1/3, SO\(_2\) 1/2.
3. \(M_{avg} = 40/6 + 64/6 + 32 = 49.33\) g/mol.
4. Density = 49.33/22.4 = 2.20 g/L at STP.

Teacher's Note:
1. Average molar mass is the mole-fraction weighted sum.
2. One mole occupies 22.4 L at STP.

Question 8: A 2 L flask of N\(_2\) at \(20^\circ\)C and 70 cm Hg is connected to a 3 L flask of O\(_2\) at the same temperature and 100 cm Hg. The final pressure and mole percent of N\(_2\) are:
(a) 100 cm Hg, 31.8%
(b) 88 cm Hg, 31.8%
(c) 88 cm Hg, 68.2%
(d) 85 cm Hg, 40%
Show Answer & Explanation

Answer: (b) 88 cm Hg, 31.8%

Explanation:
1. Total \(PV\) is conserved at constant T: \(70 \times 2 + 100 \times 3 = 440\).
2. Final volume 5 L, so \(P = 88\) cm Hg.
3. Mole percent is proportional to \(PV\): N\(_2\) = 140/440 = 31.8%.

Teacher's Note:
1. At constant T, PV is proportional to moles.
2. The pressure of a gas mixture is the sum of the partial pressures after expansion.

Question 9: Two equal flasks joined by a narrow tube each hold 0.60 mol of O\(_2\) at 0.5 atm and 300 K. One flask is placed in a thermostat at 600 K. The final pressure and the moles of O\(_2\) in the 300 K flask are:
(a) 0.5 atm, 0.6 mol
(b) 0.667 atm, 0.4 mol
(c) 0.667 atm, 0.8 mol
(d) 1.0 atm, 0.8 mol
Show Answer & Explanation

Answer: (c) 0.667 atm, 0.8 mol

Explanation:
1. Total moles = 1.2, and pressure is equal in both flasks.
2. So \(n_1 \times 300 = n_2 \times 600\), giving \(n_1 = 2n_2\): \(n_1 = 0.8\), \(n_2 = 0.4\).
3. \(V = nRT/P = 0.6 \times R \times 300/0.5 = 0.36 R \times 300\); then \(P = 0.8 R \times 300/V = 0.667\) atm.

Teacher's Note:
1. Equal pressure means n is inversely proportional to T for equal volumes.
2. The cooler flask holds more gas.

Question 10: The pressure of oxygen in a vessel falls from 2000 to 1500 torr in 40 min by leaking into vacuum. The same vessel with another gas takes 80 min for the same fall. The molar mass of the other gas is:
(a) 128 g/mol
(b) 8 g/mol
(c) 256 g/mol
(d) 64 g/mol
Show Answer & Explanation

Answer: (a) 128 g/mol

Explanation:
1. The leak time is proportional to \(\sqrt{M}\).
2. \(80/40 = \sqrt{M/32}\), so \(M = 32 \times 4 = 128\) g/mol.

Teacher's Note:
1. Doubling the time means the molar mass is four times larger.
2. Graham's law holds for effusion into vacuum.

Question 11: Nitrogen and an unknown gas effuse for 10 minutes each from identical holes at \(0^\circ\)C into a 4 L vessel. The pressure is 2.8 atm and the mixture contains 0.4 mol N\(_2\). The molar mass of the unknown gas is about (R = 0.082 L atm K\(^{-1}\) mol\(^{-1}\)):
(a) 7 g/mol
(b) 448 g/mol
(c) 112 g/mol
(d) 28 g/mol
Show Answer & Explanation

Answer: (b) 448 g/mol

Explanation:
1. \(n_{total} = PV/RT = 2.8 \times 4/(0.082 \times 273) = 0.5\) mol, so the unknown gas is 0.1 mol.
2. Moles effused in equal time are inversely proportional to \(\sqrt{M}\): \(0.4/0.1 = \sqrt{M/28}\).
3. \(M = 16 \times 28 = 448\) g/mol.

Teacher's Note:
1. Equal time and equal hole size mean the moles effused give the rate ratio directly.
2. The ratio 4 means M is 16 times the molar mass of N\(_2\).

Question 12: The rms speed of molecules, one third of which have rms speed \(10^4\), one third \(2 \times 10^4\) and one third \(3 \times 10^4\) cm/s, is:
(a) \(1.4 \times 10^4\) cm/s
(b) \(2 \times 10^4\) cm/s
(c) \(2.5 \times 10^4\) cm/s
(d) \(2.16 \times 10^4\) cm/s
Show Answer & Explanation

Answer: (d) \(2.16 \times 10^4\) cm/s

Explanation:
1. Mean of the squares: \((1 + 4 + 9)/3 \times 10^8 = 4.67 \times 10^8\).
2. Square root: \(2.16 \times 10^4\) cm/s.

Teacher's Note:
1. Average the squares first, then take the root.
2. The rms speed is larger than the simple mean 2 \(\times 10^4\).

Question 13: The rms speed of a gas at 27 K and 1.5 bar is \(1 \times 10^4\) cm/s. If both temperature and pressure are raised three times, the new rms speed is:
(a) \(5.2 \times 10^4\) cm/s
(b) \(1 \times 10^4\) cm/s
(c) \(1.73 \times 10^4\) cm/s
(d) \(3 \times 10^4\) cm/s
Show Answer & Explanation

Answer: (c) \(1.73 \times 10^4\) cm/s

Explanation:
1. The rms speed depends only on \(\sqrt{T/M}\), not on pressure.
2. New speed = \(10^4 \times \sqrt{3} = 1.73 \times 10^4\) cm/s.

Teacher's Note:
1. Pressure has no direct effect on the rms speed of an ideal gas.
2. Speed scales with the square root of absolute temperature.

Question 14: At what temperature (in \(^\circ\)C) is the most probable speed of CO\(_2\) twice that at \(127^\circ\)C?
(a) 800
(b) 527
(c) 1327
(d) 1600
Show Answer & Explanation

Answer: (c) 1327

Explanation:
1. \(u_{mp} \propto \sqrt{T}\), so doubling the speed needs 4 times the kelvin temperature.
2. \(T = 4 \times 400 = 1600\) K = \(1327^\circ\)C.

Teacher's Note:
1. Convert to kelvin before scaling.
2. Do not report 1600 K as \(^\circ\)C.

Question 15: Hydrogen molecules have the same rms speed as nitrogen molecules at \(35^\circ\)C at a temperature of:
(a) 308 K
(b) 44 K
(c) 4312 K
(d) 22 K
Show Answer & Explanation

Answer: (d) 22 K

Explanation:
1. \(u_{rms} \propto \sqrt{T/M}\), so \(T_{H_2}/2 = 308/28\).
2. \(T_{H_2} = 308 \times 2/28 = 22\) K.

Teacher's Note:
1. The lighter gas needs a lower temperature to match the speed.
2. \(35^\circ\)C = 308 K.

Question 16: 1 L of a CO and CO\(_2\) mixture is passed over red hot charcoal and the volume becomes 1.6 L (same conditions). The volume of CO in the original mixture is:
(a) 0.4 L
(b) 0.6 L
(c) 0.3 L
(d) 0.8 L
Show Answer & Explanation

Answer: (a) 0.4 L

Explanation:
1. \(\ce{CO2 + C -> 2CO}\): each volume of CO\(_2\) gives 2 volumes of CO, an increase of 1 volume.
2. Increase = 0.6 L, so CO\(_2\) = 0.6 L and CO = 0.4 L.

Teacher's Note:
1. The volume change equals the volume of CO\(_2\) converted.
2. CO + CO\(_2\) = 1 L.

Question 17: When 100 mL of an O\(_2\)-O\(_3\) mixture is passed through turpentine, the volume falls by 20 mL. If the 100 mL mixture is heated instead, the volume increases by:
(a) 10 mL
(b) 5 mL
(c) 20 mL
(d) 30 mL
Show Answer & Explanation

Answer: (a) 10 mL

Explanation:
1. Turpentine absorbs ozone, so there is 20 mL of O\(_3\).
2. \(\ce{2O3 -> 3O2}\): 20 mL O\(_3\) gives 30 mL O\(_2\).
3. Increase = 30 - 20 = 10 mL.

Teacher's Note:
1. Decomposition of ozone increases the volume by half of the ozone volume.

Question 18: 60 mL of a mixture of N\(_2\)O and NO is exploded with excess hydrogen and 38 mL of N\(_2\) forms. The volumes of N\(_2\)O and NO are:
(a) 22 mL and 38 mL
(b) 16 mL and 44 mL
(c) 44 mL and 16 mL
(d) 30 mL and 30 mL
Show Answer & Explanation

Answer: (b) 16 mL and 44 mL

Explanation:
1. Let N\(_2\)O = x, NO = y.
2. \(\ce{N2O + H2 -> N2 + H2O}\) gives x mL N\(_2\); \(\ce{2NO + 2H2 -> N2 + 2H2O}\) gives y/2.
3. \(x + y = 60\), \(x + y/2 = 38\) gives \(y = 44\), \(x = 16\).

Teacher's Note:
1. Write the balanced equation for each oxide.
2. NO gives half its volume of N\(_2\).

Question 19: A mixture of HCOOH and \(\ce{H2C2O4}\) is heated with conc. \(\ce{H2SO4}\). The gas evolved falls by one-sixth in volume when passed through KOH. The molar ratio of HCOOH to \(\ce{H2C2O4}\) is:
(a) 1 : 1
(b) 4 : 1
(c) 2 : 1
(d) 1 : 4
Show Answer & Explanation

Answer: (b) 4 : 1

Explanation:
1. HCOOH gives 1 CO; \(\ce{H2C2O4}\) gives 1 CO + 1 CO\(_2\).
2. With a mol HCOOH and b mol oxalic acid: gas = a + 2b, CO\(_2\) = b.
3. \(b = (a + 2b)/6\), so \(a = 4b\).

Teacher's Note:
1. KOH absorbs only CO\(_2\).
2. H\(_2\)SO\(_4\) acts as a dehydrating agent here.

Question 20: A gaseous hydrocarbon (1.12 L at NTP) on complete combustion gives 2.2 g CO\(_2\) and 1.8 g H\(_2\)O. The hydrocarbon is:
(a) C\(_3\)H\(_8\)
(b) C\(_2\)H\(_4\)
(c) C\(_2\)H\(_6\)
(d) CH\(_4\)
Show Answer & Explanation

Answer: (d) CH\(_4\)

Explanation:
1. Hydrocarbon = 1.12/22.4 = 0.05 mol.
2. CO\(_2\) = 2.2/44 = 0.05 mol, so 1 C per molecule.
3. H\(_2\)O = 0.1 mol, so 4 H per molecule. The formula is CH\(_4\).

Teacher's Note:
1. Moles of each product divided by moles of hydrocarbon give the atoms per molecule.

Question 21: What height (in cm) of liquid gallium (density 6.095 g/mL) is supported when the mercury barometer reads 740 torr (density of Hg = 13.6 g/mL)?
(a) 165
(b) 285
(c) 322
(d) 210
Show Answer & Explanation

Answer: (a) 165

Explanation:
1. Pressure \(= h\rho g\) is the same for both liquids.
2. \(h_{Ga} = 74.0 \times 13.6/6.095 = 165\) cm.

Teacher's Note:
1. Lighter liquids rise higher for the same pressure.

Question 22: A gas is heated from \(0^\circ\)C to \(100^\circ\)C at 1.0 atm. If the initial volume is 10.0 L, the final volume is:
(a) 10.00 L
(b) 13.66 L
(c) 20.00 L
(d) 7.32 L
Show Answer & Explanation

Answer: (b) 13.66 L

Explanation:
1. \(V_2 = V_1 T_2/T_1 = 10.0 \times 373/273 = 13.66\) L.

Teacher's Note:
1. Volume does not double unless the kelvin temperature doubles.

Question 23: The pressure of a gas in a closed vessel rises by 0.4% when heated by \(1^\circ\)C. The initial temperature was:
(a) 25 K
(b) 250\(^\circ\)C
(c) 25\(^\circ\)C
(d) 250 K
Show Answer & Explanation

Answer: (d) 250 K

Explanation:
1. At constant volume, \(\Delta P/P = \Delta T/T\).
2. \(0.004 = 1/T\), so \(T = 250\) K.

Teacher's Note:
1. The fractional rise in pressure equals the fractional rise in temperature.

Question 24: A balloon of air at \(47^\circ\)C has volume 3 L. In a cooler room it becomes 2.7 L. The temperature of the room is:
(a) \(100^\circ\)C
(b) \(15^\circ\)C
(c) \(200^\circ\)C
(d) \(42^\circ\)C
Show Answer & Explanation

Answer: (b) \(15^\circ\)C

Explanation:
1. \(T_2 = 320 \times 2.7/3 = 288\) K.
2. \(288 - 273 = 15^\circ\)C.

Teacher's Note:
1. Use kelvin in Charles' law.

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