71.  Which one of the following statements is incorrect ?

A. All elements are homogeneous system

B. Compounds made up of a number of elements are heterogeneous.

C. A mixture is not always heterogeneous

D. Smoke is a heterogeneous mixture.

Answer : Option B

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72.  A balanced chemical equation is in accordance with

A. Avogadro’s law.

B. Law of constant proportions

C. Law of conservation of mass

D. Law of gaseous volumes.

Answer : Option C

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73.  The atomic weights of two elements X and Y are 20 and 40 respectively. If ‘a’ gm of X contains ‘b’ atoms, how many atoms are present in ‘2a’ gm of Y ?

A. b

B. a

C. 2b

D. a/2

Answer : Option A

Explanation :

No of moles of X =

 No of atoms of X  =

No of moles of Y =

 No of atoms of Y  =

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74.  If the components of air are ,78%, , 21% ; Ar, 0.9% and , 0.1% by volume, what will be the molecular weight of air ?

A. 28.9

B. 32.4

C. 16.4

D. 14.5

Answer : Option A

Explanation :

Mol wt of air =

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75.  Calculate the molarity of a solution obtained by mixing 50mL of 0.5M and 75 mL of 0.25M

A. 0.375M

B. 0.35M

C. 0.045M

D. 0.45M

Answer : Option B

Explanation :

→ No.of moles of 0.05L  = 0.5 x 0.05 = 0.025

→ No.of moles of 0.075L  = 0.25 x 0.075 = 0.01875

Total no. of moles = 0.025 + 0.01875 = 0.04375

Total vol = 0.05L + 0.075L = 0.125L

 Molarity =

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76.  Which of the following has the highest normality ?

A. 1M

B. 1M

C. 1M

D. 1M

Answer : Option C

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77.  In an experiment, 4 gm of oxide was reduced to 2.8 gm of the metal. If the atomic mass of the metal is 56 gm/mol, the number of oxygen atoms in the oxide is (AFMC 2010)

A. 1

B. 2

C. 3

D. 4

Answer : Option C

Explanation :

→ 1 Mol   = (2 x 56 + 16x) gm

→ Now, (2 x 56 +16x) gm of oxide = 112 gm metal

4 gm of oxide =  gm metal

→ But  = 2.8 (given)

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78.  

Match the following

Column - I    Column - II

(i) femto        (P)

(ii) yotta         (q)

(iii) giga         (r)

(iv) atto          (s)

A. i - q, ii - p, iii - r, iv - s

B. i - s, ii - q, iii - p, iv - r

C. i - q, ii - s, iii - p, iv - r

D. i - r, ii - s, iii - p, iv - q

Answer : Option C

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79.  The total number of atoms of all elements present in mole of ammonium dichromate is

A. 19

B.

C.

D.

Answer : Option C

Explanation :

→ Molecular formula of ammonium dichromate is

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80.  0.32 gm of a metal on treatment with an acid gave 112 mL of hydrogen at STP. Calculate the equivalent weight of the metal

A. 58

B. 32

C. 11.2

D. 24

Answer : Option B

Explanation :

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