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NEET NEET Questions — Free MCQ Practice

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A mixture contains Na₂CO₃ and NaHCO₃ and wt. of mixture is 10gm. Mixture on heating liberates 56ml of CO₂ at S.T.P, wt of Na₂CO₃ in mixture is

A.

9.58 gm

B.

5 gm

C.

7.5 gm

D.

8 gm

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Explanation

Only NaHCO₃evolve CO₂ gas. Na₂CO₃ does not evolve CO₂ gas even on red heating.

So, the only reaction happening is:

2NaHCO₃ -> Na₂CO₃ + CO₂ + H₂O

Weight:

2(23+1+12+16*3) -> (2*23+12+16*3) + (12+16*2) + (1*2+16)

168g -> 102g + 44g + 18g

Also, CO₂mass as per 56ml volume obtained at STP = 56/(22.4 * 1000) * (44) = 0.11g

So, NaHCO₃weight = 0.11/44*168 = 0.42g (using ratio of weights from equation)

Therefore, original Na₂CO₃ weight in the original mixture = 10 - 0.42 = 9.58g

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How many moles of KMnO₄ are needed a mixture of 1 mole of each FeSO₄ & FeC₂O₄ in acidic medium

A.

$\frac{4}{5}$

B.

$\frac{5}{4}$

C.

$\frac{3}{4}$

D.

$\frac{5}{3}$

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Explanation

Equivalents of KMnO₄ = equivalent of FeSO₄ + equivalent of FeC₂O₄

x × 5=1 × 1 + 1 × 3

x = $\frac{4}{5}$ mole

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The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to, an excess of potassium iodide solution liberating iodine according to the following equation 2Cu² (aq) + 4I^– (aq) $\overset{}{⇌}$ 2CuI(s) + I₂(aq) The iodine liberated required 24.5cm³ of a 0.100 mole dm^-3 solution of sodium thiosulphate 2S₂O₃^2- (aq) + I₂(aq) $\to_{}^{}$ 2I^– (aq) + S₄O₆^2- (aq) the percentage of copper, by mass in the copper(ll) salt is. [Atomic mass of copper = 63.5]

A.

64.2

B.

51.0

C.

48.4

D.

25.5

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Explanation

From given reactions

mmoles of hypo = mmoles of iodine × 2

= mmoles of Cu²⁺ ions

= 24.5 × 0.1 mmoles

So mass of copper = 24.5 × 0.1 × 10^–3 × 63.5 gm

So % of copper = $\frac{24 .5 \times 0 .1 \times 10^{- 3} \times 63 .5}{0 .305} \times 100 % ≃ 51 .0 %$

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Oxygen contains 90% O¹⁶ and 10% O¹⁸. Its atomic mass is [KCET 1998]

A.

17.4

B.

16.2

C.

16.5

D.

17

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Explanation

Average atomic mass of oxygen $= \frac{90 \times 16 + 10 \times 18}{100} = 16 .20$

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KClO₃ on heating decomposes to KCl and O₂. The volume of O₂at STP liberated by 0.1 mole KClO₃is

A.

4.36 L

B.

3.36 L

C.

2.36 L

D.

none of these

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Explanation

On heating KClO₃ dissociates as:

$2 K C l O_{3} \overset{Δ}{\to} 2 K C l + 3 O_{2}$

2 moles 3 × 22.4 L at STP

$∵$ 2 moles of KClO₃on heating produces = 67. 2 L of O₂ at STP

$∵$ 0.1 mole of KClO₃on heating produces = $\frac{67 .2}{2} \times 0 .1 L$ = 3.36 L of O₂ at STP

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At S.T.P. the density of CCl₄ vapour in g/L will be nearest to [CBSE PMT 1988]

A.

6.84

B.

3.42

C.

10.26

D.

4.57

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Explanation

1 mole of CCl₄ vapour = 12 + 4 × 35.5 = 154 gm = 22.4 L at S.T.P.

∴ Density $= \frac{154}{22 .4} g m L^{- 1} = 6 .875 g m L^{- 1}$

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1 c.c of N₂O at NTP contains : [CBSE PMT 1988]

A.

$\frac{1 .8}{224} \times 10^{22}$ atoms

B.

$\frac{6 .02}{22400} \times 10^{23}$ molecules

C.

$\frac{1 .32}{224} \times 10^{23}$ electrons

D.

All the above

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Explanation

22400 c.c. = 6.02 × 10²³ molecules

1 c.c. of $N_{2} O = \frac{6 .02 \times 10^{23}}{22400}$ molecules

$= \frac{3 \times 6 .02 \times 10^{23}}{22400}$ atoms (Since N₂O has three atoms)

$= \frac{6 .02 \times 10^{23}}{22400} \times 22$ electron (Because number of electrons in N₂O are 22)

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The mass of carbon present in 0.5 mole of K₄[Fe(CN)₆] is

A.

1.8 g

B.

18 g

C.

3.6 g

D.

36 g

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Explanation

1 mole of K₄[Fe(CN)₆] =6 gm atoms of carbon

0.5 mole of K₄[Fe(CN)₆] = 3 gm atoms of carbon

= 3 × 12 = 36 g

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The number of moles of BaCO₃ which contains 1.5 moles of oxygen atoms is [EAMCET 1991]

A.

0.5

B.

1

C.

3

D.

6.02 ×10²³

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Explanation

$âˆµ$ 1 mole of BaCO₃contains 3 moles of oxygen atoms.

∴ $\frac{1}{2}$ mole (0.5) of BaCO₃ contains 1.5 moles of oxygen atoms.

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The oxide of a metal contains 40% by mass of oxygen. The percentage of chlorine in the chloride of the metal is [BIT Ranchi 1997]

A.

84.7

B.

74.7

C.

64.7

D.

44.7

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Explanation

% of oxygen = $\frac{8}{m + 8} \times 100 = 40$ or, $40 (m + 8) = 800$ or, m + 8 = 20or, m = 12

∴ % of chlorine = $\frac{35 .5}{m + 35 .5} \times 100$ = $\frac{35 .5}{12 + 35 .5} \times 100$ = 74.7(Where m is the atomic mass of metal)

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