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Chemistry NEET Questions — Free MCQ Practice

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All the three states H₂O, i.e., the triple point for H₂O the equilibrium,

Ice $⇌$ Water $⇌$ Vapour exist at:

A.

3.85 mm and 0.0981 $°$ C

B.

4.58 mm and 0.0098 $°$ C

C.

760 mm and 0 $°$ C

D.

none of the above

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Explanation

(b) The conditions for triple point of H₂O.

the particular temperature and pressure at which the solid, liquid, and gaseous phases of a given substance are all at equilibrium with one another.

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The beans are cooked earlier in pressure cooker, because

A.

boiling point increases with increasing pressure

B.

boiling point decreases with increasing pressure

C.

extra pressure of pressure cooker softens the beans

D.

internal energy is not lost while cooking in pressure cooker

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Explanation

(a)

The beans are cooked earlier in the pressure cooker because the boiling point increases with increasing pressure.

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Boyle's law is applicable in :

A.

isobaric process

B.

isochoric process

C.

isothermal process

D.

adiabatic process

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Explanation

(c) A constant temperature refers for isothermal process.

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Internal energy and pressure of a gas per unit volume are related as

A.

p= $\frac{2}{3}$ E

B.

p = $\frac{3}{2}$ E

C.

p= $\frac{1}{2}$ E

D.

p = 2E

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Explanation

(a) The average translational kinetic energy of a gas molecule is $\frac{1}{2}$ m $u$ ² at a temperature T. The total energy of the whole of the gas containing N molecules is

E_k = $\frac{1}{2}$ mN $u$ ² ....(i)

The kinetic gas equation is

pV = $\frac{1}{3}$ mN $u$ ² .... (ii)

pV = 2/3 x 1/2 mN $u$ ²

pV = $\frac{2}{3}$ E_k

p = $\frac{2}{3}$ E_k per unit volume

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A sample of water gas has a composition by volume of 50% H₂, 45% CO and 5% CO₂. Calculate the volume in litre at S.T.P. of water gas which on treatment with excess of stream will produce 5 litre H₂. The equation for the reaction is : CO + H₂O $\overset{}{\to}$ CO₂+ H₂

A.

4.263 Litre

B.

5.263 Litre

C.

6.263 Litre

D.

7.263 Litre

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Explanation

Let suppose we use x–litre water gas at S.T.P.

So it contain 0.5x litre H₂.

Now according to given equation

CO + H₂O $\overset{}{\to}$ CO₂+ H₂

0.45x Excess 0.05x 0.5x (45% CO , and 5% CO2)

0 –(0.05x + 0.45x) (0.5x + 0.45x)

So total volume of H₂ = 0.95x

But according to question

It is 5 litre

So 0.95x = 5

x = 5.263 litre

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When 22.4 L of H₂ (g) is mixed with 11.2 L of Cl₂ (g), each at STP, the moles of HCl(g) formed is equal to

A.

1 mole of HCl (g)

B.

2 moles of HCl (g)

C.

0.5 mole of HCl (g)

D.

1.5 mole of HCl (g)

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Explanation

The given problem is related to the concept of stiochiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction].

The limiting reagent gives the moles of product formed in the reaction.

H₂(g) + Cl₂(g) $\to$ 2HCl (g)

Initial vol 22.4 L 11.2 L 2 mol

... 22.4 L volume at STP is occupied by,

Cl₂ = 1 mole,

... 11.2 L volume will be occupied by,

Cl₂ = 1x 11.2/22.4 mole = 0.5 mol

Thus, H₂(g) + Cl₂(g) $\to$ 2HCl (g)

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Which of the following contains the greatest number of nitrogen atoms ?

A.

500 ml of 2.0 $M {NH}_{3}$

B.

One mole of ${NH}_{4} Cl$

C.

$6.02 \times 10^{23}$ molecules of ${NO}_{2}$ gas

D.

22.4 litres of $N_{2}$ gas at 0ºC and 1 atm.

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Explanation

(D). 22.4 litres of $N_{2}$ will be present in 1 mole at 1 atmospheric pressure and 273 K or 0ºC.

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The values of vander waals constant ‘a’ for the gases $O_{2}$ , $N_{2}$ , ${NH}_{3}$ and ${CH}_{4}$ are 1.36, 1.39, 4.17 and 2.253 ${lit}^{2}$ atom mol^-2 respectively. The gas which can most easily be liquefied is $-$

A.

$O_{2}$

B.

$N_{2}$

C.

${NH}_{3}$

D.

${CH}_{4}$

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Explanation

(C). More the ‘a’ value of the gas, more will be the inter molecular attraction between the gas

molecules, therefore, easier will be the liquefaction.

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Given reaction : $C (s) + H_{2} O (g) \to CO (g) + H_{2} (g)$ . Calculate the volume at STP from 48 gm of carbon and excess $H_{2} O -$

A.

179.2 lit.

B.

89.6 lit.

C.

44.8 lit.

D.

22.4 lit.

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Explanation

$(A) . 48 gm . C = \frac{48}{12} = 4 moles of C . From 4 moles of C, 4 moles of CO (g) and 4 moles of H_{2} (g) is obtained . ∴ total moles of gas produced = 8 ∴ volume of product gas mixture at STP = 8 \times 22.4 = 179.2 lit .$

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A mixture of 10 ml ${CH}_{4}, C_{2} H_{4} and C_{2} H_{2}$ has a vapour density of 11.3. Mixture contains x ml of ${CH}_{4}$ , y ml of $C_{2} H_{4}$ and z ml of $C_{2} H_{2}$ . When 30 ml of oxygen are sparked together over aqueous KOH, the volume contracts to 5.5 ml and then disappears when pyrogallol is introduced. If volumes are measured in the same conditions of pressure, temperature and humidity, value of x, y and z is–

A.

4, 3, 3

B.

1, 2, 3

C.

2, 1, 1,

D.

3, 4, 2

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Explanation

(A). Let the volume of ${CH}_{4}$ at N.T.P. = x ml

Let the volume of $C_{2} H_{4}$ at N.T.P. = y ml

Let the volume of $C_{2} H_{2}$ at N.T.P. = z ml

From question, x + y + z = 10 ............ (1)

As we know that

Weight of ${CH}_{4}$ + Weight of $C_{2} H_{4}$ + Weight of $C_{2} H_{2}$ = Weight of mixture

$\frac{16 x}{22400} + \frac{28 x}{22400} + \frac{262}{22400} = \frac{11.30}{11200} \times 10$ .....................(2)

$Now, \underset{x ml}{{CH}_{4}} (g) + \underset{2 x ml}{2 O_{2}} (g) ⇌ {CO}_{2} (g) + 2 H_{2} O (l) \underset{y ml}{C_{2} H_{4}} (g) + \underset{3 y ml}{3 O_{2}} (g) ⇌ 2 {CO}_{2} (g) + 2 H_{2} O (l) \underset{z ml}{C_{2} H_{2}} (g) + \underset{\frac{5}{2} z ml}{\frac{5}{2} O_{2}} (g) ⇌ 2 {CO}_{2} (g) + H_{2} O (l)$

Total volume of oxygen used up in the reaction $= (2 x + 3 y + \frac{5}{2} z) ml$

But from question,

Total volume of oxygen used up $= 30 - 5.5 = 24.5 ml$

$∴ 2 x + 3 y + \frac{5}{2} z = 24.5$ ................... (3)

Solving equations (1), (2) and (3), we get

x=4, y=3, z=3

$∴ {CH}_{4} = 4 ml, C_{2} H_{4} = 3 ml, C_{2} H_{2} = 3 ml$

Chemistry MCQs for NEET — Practice Questions with Answers

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