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Physics MCQs for NEET — Practice Questions with Answers

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The volume of 2.8 g of carbon monoxide at 27°C and 0.821 atm pressure is $(R = 0 .0821 l i t a t m K^{- 1} m o l^{- 1})$ [Manipal PMT 2001]

0.3 litre

1.5 litre

3 litre

30 litre

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Explanation

2.8 g CO $= \frac{2 .8}{28} = m o l = 0 .1 m o l$

PV = nRT

or $V = \frac{n R T}{P} = \frac{0 .1 \times 0 .0821 \times 300}{0 .821} = 3 l i t r e$

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The density of methane at 2.0 atmosphere pressure and 27°C is [BHU 1994]

0.13 gL^–1

0.26 gL^–1

1.30 gL^–1

2.60 gL^–1

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Explanation

$d = \frac{P M}{R T} = \frac{2 \times 16}{0 .082 \times 300} = 1 .30 g L^{- 1}$

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The volume of 0.0168 mol of O₂ obtained by decomposition of KClO₃ and collected by displacement of water is 428 ml at a pressure of 754 mm Hg at 25°C. The pressure of water vapour at 25°C is [UPSEAT 1996]

18 mm Hg

20 mm Hg

22 mm Hg

24 mm Hg

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Explanation

Volume of 0.0168 mol of O₂ at STP $= 0 .0168 \times 22400 c c = 376 .3 c c$

$V_{1} = 376 .3 c c$ , $P_{1} = 760 m m$ , $T_{1} = 273 K$

$V_{2} = 428 c c$ , $P_{2} = ?$ , $T_{2} = 298 K$

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ gives $P_{2} = 730 m m$ (approx.)

∴ Pressure of water vapour = 754 – 730 = 24 mm Hg

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What will be the partial pressure of H₂ in a flask containing 2 g of H₂, 14 g of N₂ and 16 g of O₂[Assam JET 1992]

1/2 the total pressure

1/3 the total pressure

1/4 the total pressure

1/16 the total pressure

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Explanation

$n (H_{2}) = \frac{2}{2} = 1, n (N_{2}) = \frac{14}{28} = 0 .5$ ,

$n (O_{2}) = \frac{16}{32} = 0 .5, p (H_{2}) = \frac{1}{1 + 0 .5 + 0 .5} p = \frac{1}{2} p$

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Equal weights of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is

1/3

1/2

2/3

$1 / 3 \times 273 / 298$

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Explanation

$n (C H_{4}) = \frac{w}{16} = 1, n (O_{2}) = \frac{w}{32}$

$p (O_{2}) = \frac{w / 32}{w / 16 + w / 32} = \frac{1}{2 + 1} = \frac{1}{3}$

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In a flask of volume V litres, 0.2 mol of oxygen, 0.4 mol of nitrogen, 0.1 mol of ammonia and 0.3 mol of helium are enclosed at $27^{o} C$ . If the total pressure exerted by these non-reating gases is one atmosphere, the partial pressure exerted by nitrogen is [Kerala MEE 2001]

1 atm

0.1 atm

0.2 atm

0.4 atm

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Explanation

$P_{N_{2}} = Mol. fraction of N_{2} \times Total perssure = \frac{0.4}{0.2 + 0.4 + 0.1 + 0.3} \times 1 a t m = 0 .4 a t m$ .

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Equal weights of ethane and hydrogen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by hydrogen is [IIT 1993]

1 : 2

1 : 1

1 : 16

15 : 16

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Explanation

$n (C_{2} H_{6}) = \frac{w}{30}, n (H_{2}) = \frac{w}{2}$

$p (H_{2}) = \frac{w / 2}{w / 2 + w / 30} = \frac{1}{1 + \frac{1}{15}} = \frac{15}{16}$

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A gaseous mixture contains 56 g of N₂, 44 g of CO₂ and 16 g of CH₄. The total pressure of the mixture is 720 mm Hg. The partial pressure of CH₄ is [IIT 1993]

180 mm

360 mm

540 mm

720 mm

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Explanation

$p (C H_{4}) = \frac{16 / 16}{5 / 28 + 44 / 44 + 16 / 16} \times 720 = \frac{1}{2 + 1 + 1} \times 720 = \frac{1}{4} \times 720 = 180 m m$ .

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The time taken for a certain volume of a gas ‘X’ to diffuse through a small hole is 2 minutes. It takes 5.65 minutes for oxygen to diffuse under the similar conditions. The molecular weight of ‘X’ is [NCERT 1990]

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Explanation

$\frac{r_{X}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{X}}}$

$\frac{v / 2}{v / 5 .65} = \sqrt{\frac{32}{M_{X}}}, \frac{5 .65}{2} = \sqrt{\frac{32}{M_{X}}}, M_{X} = 4$

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The rate of diffusion of methane at a given temperature is twice that of gas X. The molecular weight of X is

64.0

32.0

4.0

8.0

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Explanation

$\frac{r_{C H_{4}}}{r_{X}} = \sqrt{\frac{M_{X}}{M_{C H_{4}}}} \Rightarrow 2 = \sqrt{\frac{M_{X}}{16}} \Rightarrow M_{X} = 64$ .

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