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Chemistry NEET Questions — Free MCQ Practice

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Assertion : The difference in the boiling points of equimolar solution of HCI and HF decreases as their

molarity is decreased.

Reason : The extent of dissociation decreases steadily with increasing dilution.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

(C) Extent of dissociated increases steadily with increasing dilution.

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Assertion : When 'a' mL of a 0.1 molal urea solution is mixed with another 'b' mL of 0.1 molal glucose

solution, the boiling point of the solution is no different from the boiling points of the

samples prior to mixing but if 'a' mL of 0.1 molal urea is mixed with 'b' mL of 0.1 molal HF

the boiling point of the mixture is different from the boiling points of the separate

samples.

Reason : HF is an electrolyte (weak) whereas glucose is a non electrolyte.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

(A) As degree of dissociation of HF gets changed on dilution.

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An electrochemical cell is shown below Pt, H₂(1 atm)| HCI (0.1 M)CH₃COOH (0.1 M)|

H₂(1 atm), Pt The EMF of the cell will not be zero, because

A.

EMF depends on molarities of acids used

B.

pH of 0.1 M HCl and 0.1 M CH₃COOH is not same

C.

the temperature is constant

D.

acids used in two compartments are different

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Explanation

(b) The EMF of the cell will not be zero because concentration of H⁺ ions in two electrolytic solutions is different. Mean HCl is strong acid where, acetic acid is weak acid and gives different pH.

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Saturated solution of KNO₃ is used to make 'salt-bridge' because:

A.

velocity of K⁺ is greater than that of ${NO}_{3}^{-}$

B.

velocity of ${NO}_{3}^{-}$ is greater than that of K⁺

C.

Velocities of both K⁺ and ${NO}_{3}^{-}$ are nearly the same

D.

KNO₃ is highly soluble in water

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Explanation

(c) The salt bridge possesses, the electrolyte having nearly same ionic mobilities of its cation and anion.

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A current is passed through two voltameters connected in series. The first voltmeter connected in series. The first voltmeter contains XSO₄(aq) while the second voltmeter contains Y₂SO₄(aq). The relative atomic masses of X and Y are in the ratio of 2:1. The ration of the mass of X liberated to the mass of Y liberated is:

A.

1:1

B.

1:2

C.

2:1

D.

none of these

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Explanation

(a) Equal eqivalents of each are liberated.

Eq. of X=Eq. of Y

$\frac{m_{1}}{\frac{2 M}{2}} = \frac{m_{2}}{\frac{M}{1}} ∴ m_{1} = m_{2}$

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The mass of silver(eq. mass = 108) displaced by that quantity of current which displaced 5600 mL of hydrogen at STP is:

A.

54 g

B.

108 g

C.

5.4 g

D.

none of these

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Explanation

Concept used: Faradays second law of electrolysis

Solve: Acc. to Faraday's seconf law

$\frac{w_{1}}{e q w t_{1}} = \frac{w_{2}}{e q w t_{2}}$

here since gas is involved se we can directly use

$\frac{w_{a g}}{e q_{a g}} = \frac{v o l . o f g a s a t N T P}{v o l . o f 1 g m e q .} \frac{w_{1}}{108} = \frac{5600}{\frac{22400}{2}} \frac{w_{1}}{108} = \frac{5600}{11200}$

$\frac{w}{108} = \frac{1}{2}, w = \frac{108}{2} = 54 g$

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A silver cup is plated with silver by passing 965 coulomb of electricity. The amount of Ag deposited is:

A.

1.08 g

B.

1.0002 g

C.

9.89 g

D.

107.89 g

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Explanation

Concept used: Faraday's first law.

Given : Q= 965 coulomb

wt of sub deposited is given by

w = 2 x Q (Faraday's first law)

$w = \frac{E}{96500} \times Q$

Equivalent wt of metal E= $\frac{molecular wt}{n - factor}$

$= w = \frac{\frac{M}{n - factor}}{96500} \times Q = \frac{\frac{108}{1}}{96500} \times 965 = \frac{108}{100} = 1.08 g$

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At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 Ω^-1cm²mol^-1 and at infinite dilution its molar conductance is 238 Ω^-1cm²mol^-1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is

A.

2.080 %

B.

20.800 %

C.

4.008 %

D.

40.800 %

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Explanation

(C) Given, molar conductance at 0.1 M concentration,

$λ_{c}$ = 9.54Ω^-1cm²mol^-1

Molar conductance at infinite dilution,

${λ_{c}}^{\infty}$ = 238 Ω^-1cm²mol^-1.

We know that,

degree of ionisation,

$α$ = $λ_{c}$ / ${λ_{c}}^{\infty}$

=(9.54/238) x 100 = 4.008%

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Which is the correct representation for Nernst equation ?

A.

$E_{RP} = E_{RP}^{\circ} + \frac{0.059}{n} \log \frac{[oxidant]}{[reductant]}$

B.

$E_{OP} = E_{OP}^{\circ} - \frac{0.059}{n} \log \frac{[oxidant]}{[reductant]}$

C.

$E_{OP} = E_{OP}^{\circ} + \frac{0.059}{n} \log \frac{[reductant]}{[oxidant]}$

D.

All of the above

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Explanation

(d)

The Nernst equation is an important relation which is used to determine reaction equilibrium constants and concentration potentials

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When a copper wire is immersed in a solution of AgNO₃, the colour of solution becomes blue because copper:

A.

forms a soluble complex with AgNO₃

B.

is oxidised to Cu²⁺

C.

is reduced to Cu^2-

D.

splits up into atoic form and dissolves

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Explanation

(b) Cu is above Ag in electrochemical series and thus,

$Cu + 2 {Ag}^{+} \to {Cu}^{2 +} + 2 Ag reaction occurs .$

Chemistry MCQs for NEET — Practice Questions with Answers

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