Blog Log in Get Started Free

Chemistry MCQs for NEET — Practice Questions with Answers

Practice free Chemistry NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

When 36 g of a solute having the empirical formula CH₂O is dissolved in 1.2 kg of water, the solution freezes at
-0.93°C. What is the molecular formula of solute (K_f= 1.86 kg K mol^-1)

C₂H₄O

C₂H₂O₂

C₂H₄O₃

C₂H₄O₂

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

ΔT_f= i K_f m

0.93= 1 x 1.86 x 36/M ; M=60
1.2

● ● ● ● ●

A solution containing 0.03659 g/ml of HCl and a solution containing 0.04509 g/ml of acetic acid Then:

N_HCl is more

N_{CH3COOH is more}

Both have same normality

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

N_HCl= $\frac{0.03659 \times 1000}{36.5} = 1.002 N$

$N_{{CH}_{3} COOH} = \frac{0.04509 \times 1000}{60} = 0.7515 N$

● ● ● ● ●

The relative lowering of vapour pressure caused by dissolving 71.3 g of a substance in 1000 g of water is 7.13 X 10^-3. The molecular mass of the substance is :-

180

1.8

360

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$d_{sol .} = \frac{mass}{volume} N = \frac{W}{E} \times \frac{1}{v (L)} V_{sol .} = \frac{100}{0.6} ml N = \frac{35}{35} \times \frac{1000 \times 0.6}{100} = 6$

● ● ● ● ●

Total vapour pressure of mixture of 1mol $A (P_{A}^{0} = 150 torr)$ and 2 mol $B (P_{B}^{0} = 240 torr)$ is 200 torr. In this case:

there is positive deviation from Raoult's law

there is negative deviation from Raoult's law

there is no deviation from Raoult's law

none of those

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The total vapor pressure of the mixture is less than the sum of the vapor pressures of the pure components (150 torr for A and 240 torr for 2 mol of B), which indicates a negative deviation from Raoult's law. This suggests that there are attractive intermolecular forces between A and B molecules, leading to a lower vapor pressure than expected.

● ● ● ● ●

When 0.1 m CH₃COOH is present in a solvent it shows elevation in boiling point of 0.75 $° C$ . Acid dissociation constant will be -(K_b=5 K Kg mol^-1) (1M =1m)

5 X 10^-2

5 X 10^-3

10^-2

10^-3

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

${ΔT}_{b} = i K_{b} m; 0.75 = i \times 5 \times 0.1 i = 1.5; α = \frac{1.5 - 1}{2 - 1} = 0.5 K_{a} = \frac{α^{2} C}{1 - α} = \frac{{(0.5)}^{2} \times 0.1}{1 - 0.5} = 5 \times 10^{- 2}$

● ● ● ● ●

Which of the following shows positive deviation from Raoult's law:

(i) Chloroform and acetone

(ii) Carbon disulphide and acetone

(iii) Ethanol and Acetone

(iv) Phenol and Aniline

i and ii

ii and iii

i and iV

ii and iv

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

A non-volatile solute is dissolved in a water. If its degree of association is 50%. What will be its freezing point when boiling point is 100.52 $° C$ . [K_b=0.52 K kg mol^-1, K_f=1.86K kg mol^-1]

-0.52 K

-0.52 $° C$

-1.86 $° C$

275.01 K

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

For a non-volatile solute, the freezing point depression (Î”Tf) is given by Î”Tf = Kf * m, where Kf is the cryoscopic constant and m is the molality of the solution. Given that the degree of association is 50%, the van't Hoff factor (i) is 2. Substituting the values, Î”Tf = 1.86 * (100.52/100) * 2 = 3.72Â°C. Therefore, the freezing point will be 100.52 - 3.72 = 96.8Â°C or -1.86Â°C.

● ● ● ● ●

When 36 g of a solute having the emperical formula CH₂O is dissolved in 1.2 kg of water, the solution freezes at -0.93 $° C$ . What is the molecular formula of solute (K_f=1.86 kg K mol^-1)

C₂H₄O

C₂H₂O₂

C₂H₄O₃

C₂H₄O₂

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$∆ T_{f} = {ik}_{f} m 0.93 = 1 \times 1.86 \times \frac{36}{1.2 M}; M = 60$

m is 60 gram , so 4th is right ans

● ● ● ● ●

How much oxygen is dissolved in 100ml water at 298K if partial pressure of oxygen is 0.5 atm and K_H=1.4 X 10^-3 M/atm ?

22.4 mg

22.4 g

2.24 g

2.24 mg

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

According to henry's law

S=K_H X p (S=conc. of O₂ dissolved)

$S = 1.4 \times 10^{- 3} \times 0.5 = 7 \times 10^{- 4} mol / L = \frac{w / M_{w}}{V_{L}} \Rightarrow w = 7 \times 10^{- 4} \times M_{w} \times V_{L} = 7 \times 10^{- 4} \times 32 \times 0.1 = 22.4 \times 10^{- 4} g = 2.24 mg$

● ● ● ● ●

A solute x when dissolved in solvent associates to form a pentamer. The value of van't hoff factor (i) for the solute will be:-

0.5

0.2

0.1

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$5 x \to x_{5} DOA, \infty \frac{1 - i}{1 - \frac{1}{n}} \Rightarrow 1 = \frac{1 - i}{1 - \frac{1}{5}} \Rightarrow \frac{4}{5} = 1 - i \Rightarrow i = 1 - \frac{4}{5} = \frac{1}{5} = 0.2$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.