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

● ● ● ● ●

Three solutions are prepared by adding 'w' gm of 'A' into 1kg of water, gm of '13' into another 1 kg of water and 'w' gm of 'C' in another 1 kg of water (A, B, C are non electrolytic). Dry air is passed from these solutions in sequence (A → B → C). The loss in weight of solution A was found to be 2gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is :

MA: M_B : Mc = 4 : 3 : 5

M_A : M_B: M_c = $\frac{1}{4} : \frac{1}{3} : \frac{1}{5}$

Mc > MA > MB

MB > M_A> M_c

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

Explanation

The loss in weight should be proportional to vapour pressure above that solution:

So, $P_{S_{A}} \propto$ 2gm

$P_{S_{B}} \propto$ 1.5gm

$P_{S_{C}} \propto$ 2.5gm

So, maximum vapour pressure is above C solution hence, it is having minimum lower and hence minimum mole fraction (hence minimum number of moles of solute) So max, molar mass of substance.

● ● ● ● ●

The molar heat of vapourization of toluene is AHv. If its vapour pressure at 315 K is 60 torr & that at 356K is300 torr then AHv = ? (log 2 = 0.3)

37.5 kJ/mole

3.75 kJ/mole

37.5 J/mol

3.75 J/mole

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

Explanation

The molar heat of vaporization (Î”Hv) can be calculated using the Clausius-Clapeyron equation: ln(P2/P1) = -(Î”Hv/R) * ((1/T2) - (1/T1)). Substituting the given values and using log(5) = 0.3 * log(2), we get: Î”Hv = (8.314 * 0.3 * (356 - 315)) / (1/356 - 1/315) = 37.5 kJ/mol.

● ● ● ● ●

Relative decrease in vapour pressure of an aqueous solution containing 2 moles [Cu(NH₃)₃Cl) Cl in 3 moles H₂O is 0.50. On reaction with AgNO₃, this solution will form (assuming no change in degree of ionisation of substance on adding AgNO₃)

1 mol AgCl

0.25 mol AgCl

0.5 mol AgCl

0.40 mol AgCl

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

Explanation

$\frac{Δ P}{P} X_{solute} = i \times \frac{2}{5} = \frac{1}{2}$

so, i = 1.25

⇒ Hence degree of dissociation = $\frac{1}{4}$

so,moles of Cl^– ions 2 × $\frac{1}{4}$ = $\frac{1}{2}$ moles

so, moles of AgCl ppt = $\frac{1}{2}$ moles = 0.5 mol AgCl

● ● ● ● ●

Which of the following has been arranged in order of decreasing freezing point?

0.05 M KNO₃> 0.04 M CaCl₂> 0.140 M sugar > 0.075 M CuSO4

0.04 M BaCl₂> 0.140 M sucrose > 0.075 M CuSO₄> 0.05 M KNO₃

0.075 MCuSO₄> 0.140 M sucrose > 0.04 M BaCl₂> 0.05 M KNO₃

0.075 M CuSO₄> 0.05 M NaNO₃> 0.140 M sucrose > 0.04 M BaCl₂

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

Explanation

Higher freezing point ⇒ lesser ΔT_f

⇒ lesser molality

⇒ lesser number of particles

● ● ● ● ●

An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to the second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of C, P_c° = 0.8 atm. Calculate the vapour pressures of pure A and pure B.

P_A° = 1.4 atm, P_B° = 0.7 atm

PA = 1.2 atm, P_B° = 0.6 atm

PA = 1.4 atm, P_B° = 0.6 atm

P_A° = 0.6 atm, P_B° = 1.4 atm

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

Explanation

$\frac{P_{A}^{0}}{2} + \frac{P_{B}^{0}}{2}$ = 1 atm $\Rightarrow$ P_A° + P_B° = 2 atm

$\frac{P_{A}^{0}}{4} + \frac{{3P}_{B}^{0}}{4}$ > 1 atmrArrP_A° + 3P_B° > 4 atm

and $\frac{P_{A}^{0}}{8} + \frac{{3P}_{B}^{0}}{8}$ + $\frac{{4P}_{C}^{0}}{8}$ = 1 atmrArrP_A° + 3P_B° + 4P_C° = 8 atm

So P_A° + 3P_B° = (8 – 4 × 0.8 ) atm = 4.8 atm

Hence P_B° = 1.4 atm

P_B° = 0.6 atm

● ● ● ● ●

A sample of air is saturated with benzene (vapor pressure = 100 mm Hg at 298 K) at 298K, 750mm Hgpressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the system is

2250 torr

2150 torr

2050 torr

1950 torr

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

Explanation

Pressure of air = 750 – 100 = 650 mm of Hg

on compressing P_f = 650 × 3 mm of Hg = 1950 mm of Hg

so P_T = (1950 + 100) = 2050 mm of Hg

● ● ● ● ●

Available solutions are 1L of 0.1 M NaCl and 2L of 0.2 M CaCl₂. Using only these two solutions what maximum volume of a solution can be prepared having [Cl^–] = 0.34M exactly. Both electrolytes are strong

2.5 L

2.4 L

2.3 L

None of these

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

Explanation

Let volumes taken to be ‘x’ & ‘y’ litres,

so $\frac{0 .1 x + 0 .4 y}{x + y}$ = 0.34&Vg = (x + y) (to be maximised)

so y = 4x so for maximum volume

y = 2L & x = $\frac{1}{2}$ L.

● ● ● ● ●

Calculate the osmotic pressure of the solution prepared in the abovequestionT = 300 K, (R = 0.082 L atm mol^-1K^-1)

10.8 atm

12.8 atm

5.6 atm

None of these

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

Explanation

π = CRT

C = (0.34 + $\frac{0 .1 \times 0 .5}{2 .5} + \frac{0 .2 \times 2}{2 .5})$ = 0.34 + 0.02 + 0.16 = 0.52

so π = 0.52 × 0.082 × 300 atm = 12.792 atm

● ● ● ● ●

Consider equimolal aqueous solutions of NaHSO₄ and NaCl with AT_b and Ar_b as their respective boiling point elevations. The value of $\lim_{x \to 0} \frac{Δ T_{b}}{Δ T_{b}^{'}}$ will be :

1.5

3.5

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

Explanation

As m → 0, NaHSO₄ will generate three particles while NaCl will generate only two particles.

● ● ● ● ●

The vapor pressures of benzene, toluene and a xylene are 75 Torr, 22 Torrand 10 Torr at 20°C. Which of the following is not a possible value of the vapor pressure of an equimolar binary/ternary solution of these at 20°C ? Assume all form ideal solution with each other.

48 $\frac{1}{2}$

$35 \frac{2}{3}$

53 $\frac{1}{2}$

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

Explanation

P_T = $\frac{1}{2}$ (75 + 22) = 48.5 torr

P_T = $\frac{1}{2}$ (75 + 10) = 42.5 torr

P_T = $\frac{1}{2}$ (22 + 10) = 16 torr

P_T = $\frac{1}{3}$ (75 + 22 + 10) = 35 $\frac{2}{3}$ torr

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.