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NEET NEET Questions — Free MCQ Practice

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Rearrange the following (I to IV) in the order of increasing masses and choose answer from (1), (2), (3) and (4) (Atomic mass : N=14, O=16 Cu = 63)

I. 1 molecule of oxygen

II. 1 atom of nitrogen

III. 1 x 10^-10g molecular weight of oxygen

IV. 1 x 10^-10g weight of copper

A.

II < I < III < IV

B.

IV < III < II < I

C.

II < III < I < IV

D.

III < IV < I < II

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Explanation

1 molecule of oxygen = 32 a.m.u=32 x 1.66x 10^-24g

1 atom of nitrogen =14 a.m.u=14 x 1.66x 10^-24g

1 x 10^-10 g molecular weight of oxygen = avg no of molecules of oxygen is 32 gram.so wt of given no is

32 x 10^-14g

1 x 10^-10g atomic weight of copper = 63.5 x 10^-14g

So, order is II < I <III <IV

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The ratio of number of atoms in 2.2g CO₂ and 1.7g NH₃ is

A.

1/2

B.

1/8

C.

3/8

D.

3/2

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Explanation

No. of atoms in 2.2 g CO₂(n₁) = $\frac{22}{44} \times 3 \times N_{A}$

No. of atoms in 1.7g NH₃(n₂) = $\frac{1.7}{17} \times 4 \times N_{A}$

$\frac{n_{1}}{n_{2}} = \frac{3}{8}$

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How many moles of FeC₂O₄ are required to reduce one mole of KMnO₄ in acidic medium?

A.

3/5

B.

1/5

C.

5/3

D.

5

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Explanation

3KMnO₄+5FeC₂O₄+24H+→5Fe³⁺+10CO2+3Mn²⁺+3K⁺+12H2O\

Ferrous is oxidized to ferric state
Carbon and hydrogen are converted to CO2 and H2O
All the metallic ions are converted to sulphates
Write the equation keeping in mind that the valency of of iron in ferric state is 3, that of K and Mn is one and two respectively as
FeC2O4 + KMnO4 + H2SO4 = K2SO4 + MnSO4 + Fe(SO4)3 + H2O + CO2
FeC2O4 is oxidized and KMnO4 is reduced. Hence the main reaction is
Fe2+ → Fe3+ (removal of 1 electron)

Mn7+ → Mn 2+ (addition of 5 electrons)

In order to balance multiply (Fe2+ → Fe3+) by 5

Now you know that there should be 5 FeC2O4 molecules. Write that and balance the rest.

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Which of the following pairs of gases contains the same number of molecules:-

A.

16g of O₂ and 14g of N₂

B.

8g of O₂ and 22g of CO₂

C.

28g of N₂ and 22g of CO₂

D.

32g of O₂ and 32g of N₂

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Explanation

Mole (n) = $\frac{W e i g h t}{M o l e c u l a r w e i g h t} = \frac{N o . o f m o l e c u l e s}{N_{A}}$

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The number of ions present in 2L of a solution of 1.6M K₄[Fe(CN)₆] is:-

A.

4.8 x 10²²

B.

4.8 x 10²³

C.

9.6 x 10²⁴

D.

9.6 x 10²²

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Explanation

No. of molecules of K₄[Fe(CN)₆] = 1.6 x 2 x 6 x 10²³

No. of ions = 5 x 1.6 x 2 x 6 x 10²³

= 9.6 x 10²⁴

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The mass of CO₂ that shall be obtained by heating 10 kg of 80% pure lime stone (CaCO₃) is:-

A.

4.4 kg

B.

6.6 kg

C.

3.52 kg

D.

8.8 kg

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Explanation

CaCO₃(S) $\overset{∆}{\to}$ CO₂(g) + CaO(s)

100 kg $\to$ 44 kg

$\frac{80}{100} \times 10 k g$ $\to$ $\frac{44}{100} \times \frac{80}{100} \times 10 = 3.52 k g$

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When 1L of CO₂ is heated with graphite,the volume of the gases collected is 1.8L. What will be the number of moles of CO produced at STP?

A.

0.0357

B.

0.0714

C.

0.0803

D.

14

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Explanation

The reaction is: C(s) + CO2(g) â†’ 2CO(g). Initially, there are 1 mol of CO2. After the reaction, the total volume is 1.8 L, which corresponds to 1.8 mol of gaseous product (assuming STP). Since 1 mol of CO2 produces 2 mol of CO, the moles of CO produced is 0.0714 mol.

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At STP the moles of oxygen atoms in 2.8 L of CO₂ gas is:-

A.

0.125

B.

0.25

C.

0.5

D.

1

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Explanation

Moles of CO₂ = $\frac{v o l u m e o f g a s}{22.4} = \frac{2.8}{22.4}$ =0.125 mol

moles of oxygen atom in given CO₂ = 0.125 x 2=0.25 mol

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Which of the following relation is incorrect:-

A.

1 mol N^-3 ions have 10 mole electrons

B.

1 mol O^-2 ions have 6.023 x 10²² electrons

C.

1 mol CH₄ has 10 mole protons

D.

1 mol water has 10 mole protons

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Explanation

1 mol N^-3 ion $\to$ 1 x 10 mol e^-

1 mol O^-2 ion $\to$ 1 x 10 mol e^-= 6.023x10^-24e^-

1 mol CH₄ $\to$ 1 x 10 mol proton

1 mol H₂O $\to$ 1 x 10 mol proton

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Mass of sucrose C₁₂H₂₂O₁₁ produced by mixing 84 gm of carbon, 12gm of hydrogen and 56L. O₂ at 1atm & 273 K according to give reaction, is C(s) + H₂(g) +O₂(g) $\to$ C₁₂H₂₂O₁₁ (s)

A.

138.5

B.

155.5

C.

172.5

D.

199.4

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Explanation

Balanced Equation:

24C + 22H₂ + 11O₂ -> 2C₁₂H₂₂O₁₁

moles for reaction:

7(84/12), 6(12/2), (56/22.4)

dividing by coefficients in balanced equation & comparing:

7/24, 6/22, (56/22.4)/11

0.29, 0.27, 0.22

Oxygen is the limiting reagent. So, weight of sucrose obtained = 0.22 * 2(coefficient for sucrose in reaction) * 342(sucrose molecular weight) = 155.5g

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