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NEET NEET Questions — Free MCQ Practice

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0.8 mole of a mixture of CO and CO₂ requires exactly 40 gram of NaOH in solution for complete conversion of all the CO₂ into Na₂CO₃. How many moles more of NaOH would it require for conversion into Na₂CO₃, if mixture (0.8 mole) is completely oxidised to CO₂?

A.

0.2

B.

0.6

C.

1

D.

1.5

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Explanation

${CO}_{2} + 2 NaOH \overset{}{\to} {Na}_{2} {CO}_{3} + H_{2} O n_{NaOH} = 1; ∴ {CO}_{2} present in mixture = 0.5 and Co present = 0.3 mole When more {CO}_{2} produced = 0.3, more NaOH requried = 0.3 \times 2 = 0.6 mole$

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The impure 6 g of $NaCl$ is dissolved in water and then treated with excess of silver nitrate solution. The weight of precipitate of silver chloride is found to be 14 g. The % purity of $NaCl$ solution would be:

A.

95%

B.

85%

C.

75%

D.

65%

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Explanation

$The reaction that takes place is NaCl + {AgNO}_{3} \overset{}{\to} AgCl ↓ + {NaNO}_{3} ∵ 143.5 g of AgCl is produced from 58.5 g NaCl ∴ 14 g of AgCl will produce from \frac{58.5 \times 14}{143.5} = 5.70 g This is the amount of NaCl in common slat; % purity = \frac{5.70}{6} \times 100 = 95 %$

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25.4 of $I_{2}$ and 14.2 g of ${Cl}_{2}$ are made to react completely to yield a mixture of ICI and ICI₃. Calculate mole of ICI and ICI₃ formed.

A.

0.5 ,02

B.

0.1, 0.1

C.

0.1 ,0.3

D.

0.3,0.4

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Explanation

$I_{2} + 2 {Cl}_{2} \overset{}{\to} ICI + {ICI}_{3} Mole before reaction \frac{25.4}{254} \frac{14.2}{71} 0 0 = 0.1 = 0.2 0 0 Mole after reaction 0 0 0.1 0.1 (Note : Mole ratio of reaction) Mole of ICI formed = 0.1 Mole of {ICI}_{3} formed = 0.1$

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The vapour density of a mixture containing ${NO}_{2}$ and $N_{2} O_{4}$ is 38.3. Calculate the mole of ${NO}_{2}$ in 100 g mixture.

A.

0.437

B.

0.347

C.

0.557

D.

0.663

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Explanation

$Given vapour density of mixture = 38.3 ∴ Molar mass of mixture = 2 \times 38.3 = 76.6 Let a g of {NO}_{2} be present in 100 g mixture, ∴ g of N_{2} O_{4} in mixture = (100 - a) Therefore, \frac{a}{46} + \frac{100 - a}{92} = \frac{100}{76.6} ∴ a = 20.10 g ∴ Mole of {NO}_{2} in mixture = 20.10 / 46 = 0.437$

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Copper forms two oxides. For the same amount of copper, twice as much oxygen was used to form first oxide than to form second one. What is the ratio of the valencies of copper in first and second oxides?

A.

2:5

B.

1:6

C.

2:1

D.

1:2

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Explanation

Let valencies of Cu in two oxides be x and y, then

I oxides is ${Cu}_{2} O_{x}$

II oxides is ${Cu}_{2} O_{y}$

In I oxide:

Equivalent of Cu = Equivalent of oxygen

$\frac{w}{A / x} = \frac{a}{8} . . . . (i)$

where w, A, x and a are mass of Cu, at. mass of Cu, valency of Cu and mass of oxygen

In II oxide:

$\frac{w}{A / y} = \frac{a}{8} . . . . (ii)$

By Eqs. (i) and (ii)

$\frac{x}{y} = \frac{2}{1}$

$∴$ Valency of Cu in I and II oxides are in the ratio 2:1

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5 mL of a gaseous hydrocarbon was exposed to 30 mL of $O_{2}$ . The resultant gas, on cooling is found to measure 25 mL of which 10 mL is absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant room temperature.

A.

CH₄

B.

C₂H₂

C.

C₃H₄

D.

C₂H₄

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Explanation

Suppose formula of hydrocarbon is $C_{a} H_{b}$

$∴ C_{a} H_{b} + [a + (b / 4)] O_{2} \overset{}{\to} {aCO}_{2} + (b / 2) H_{2} O_{(l)}$

Volume taken initially

$5 mL 30 mL - -$

Volume after reaction

$0 30 - 5 [a + b / 4] 5 a -$

$∵$ Given that, $O_{2}$ left is absorbed by pyrogallol

Also volume absorbed by NaOH = 10 mL = Volume of ${CO}_{2}$ formed

$∴ 5 a = 10 ∴ a = 2$

Total volume of gases left after reaction = 25 mL

or Volume of $O_{2}$ left + Volume of ${CO}_{2}$ formed = 25 mL

$∴$ Volume of $O_{2}$ left = 25-10 =15

$∴$ 30-5 [a+(b/4)]=15 $∴$ b=4

Thus, hydrocarbon is $C_{2} H_{4}$ .

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A sample of ${CaCO}_{3}$ and ${MgCO}_{3}$ weighed 2.21 g is ignited to constant mass of 1.152 g. What is the composition of mixture? Also calculate the volume of ${CO}_{2}$ evolved at $0 ° C$ and 76 cm of pressure.

A.

640 ml

B.

540 ml

C.

500 ml

D.

600 ml

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Explanation

$Let the mixture contains a g {CaCO}_{3} and b g {MgCO}_{3} . ∴ a + b = 2.21 . . . . . (i) On heating {CaCO}_{3} \overset{}{\to} CaO + {CO}_{2} {MgCO}_{3} \overset{}{\to} MgO + {CO}_{2} ∵ Molar of CaO = mole of {CaCO}_{3} = a / 100 ∴ Mass of CaO = (a / 100) \times 56 g Similarly, mass of MgO = (b / 84) \times 40 g Therefore, \frac{56 a}{100} + \frac{40 b}{84} = 1.152 . . . . . (ii) (i . e ., mass of residue left) Solving Eqs . (i) and (ii) a = 1.19 g; b = 1.02 g Also, mole of {CO}_{2} formed = mole of {CaCO}_{3} + mole of {MgCO}_{3} = \frac{1.19}{100} + \frac{1.02}{84} = 0.0241 ∴ Volume of {CO}_{2} at NTP = 0.0241 \times 22400 = 539.8 mL$

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The vapour density of a volume chloride of a metal is 95 and the specific heat of the metal is 0.13 cal/g. The equivalent mass of the metal will be:

A.

6

B.

12

C.

18

D.

24

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Explanation

$Molar mass of metal chloride = 95 \times 2 = 190 g {mol}^{- 1} Atomic mass of metal = \frac{6.4}{0.13} = 49.23 g {mol}^{- 1} Let the metal chloride be {MCl}_{n} Then 49.23 + n \times 35.5 = 190 ∴ n = 3.9 \approx 4; Thus exact atomic mass (a) of metal can be derived by a + 4 \times 35.5 = 190 ∴ a = 48 ∴ E = \frac{48}{4} = 12 g {eq}^{- 1}$

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Which mixture is lighter than humid air?

A.

$N_{2} + O_{2} + {SO}_{2}$

B.

$N_{2} + O_{2} + {CO}_{2}$

C.

$N_{2} + O_{2} + C_{2} H_{6}$

D.

$N_{2} + O_{2} + He$

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Explanation

Humid air $N_{2} + O_{2} + H_{2} O_{v}$ is lighter than dry air $(N_{2} + O_{2})$ as its average molar mass is less. Presence of $He$ in air $(N_{2} + O_{2})$ gives lower molar mass than humid air. $He$ has lower molar mass than $H_{2} O_{v}$ .

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$O_{3}$ is absorbed by:

A.

Pyrogallol

B.

Menthol

C.

Turpentine oil

D.

Clove oil

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Explanation

$O_{3}$ reacts with turpentine oil or oil of cinnamon.

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