1
MCQ (Single Correct Answer)

AIPMT 2004

The maximum number of molecules is present in
A
15 L of H2 gas at STP
B
5 L of N2 gas at STP
C
0.5 g of H2 gas
D
10 g of O2 gas

Explanation

At STP, 22.4 L H2

= 6.023 $$ \times $$ 1023 molecules

15 L H2 = $${{6.023 \times {{10}^{23}} \times 15} \over {22.4}}$$ = $$4.033 \times {10^{23}}$$

5 L N2 = $${{6.023 \times {{10}^{23}} \times 5} \over {22.4}}$$ = 1.344 $$ \times $$ 1023

2 g of H2 = 6.023 $$ \times $$ 1023

0.5 g H2 = $${{6.023 \times {{10}^{23}} \times 0.5} \over {2}}$$ = 1.505 $$ \times $$ 1023

32 g O2 = 6.023 $$ \times $$ 1023

10 g O2 = $${{6.023 \times {{10}^{23}} \times 10} \over {32}}$$ = 1.882 $$ \times $$ 1023
2
MCQ (Single Correct Answer)

AIPMT 2002

The percentage of C, H and N in an organic compound are 40%, 13.3% and 46.7% respectively then empirical formula is
A
C3H13N3
B
CH2N
C
CH4N
D
CH6N

Explanation

Element % Atomic
mass
Relative
no. of atoms
Simplest
ratio of atoms
1 C 40 12 $${{40} \over {12}}$$ = 3.33 $${{3.33} \over {3.3}}$$ = 1
2 H 13.3 1 $${{13.3} \over 1}$$ = 13.3 $${{13.3} \over {3.3}}$$ = 4
2 N 46.7 14 $${{46.7} \over {14}}$$ = 3.3 $${{3.3} \over {3.3}}$$ = 1


$$ \therefore $$ The empirical formula is CH4N.
3
MCQ (Single Correct Answer)

AIPMT 2002

Which has maximum molecules?
A
7 g N2
B
2 g H2
C
16 g NO2
D
16 g O2

Explanation

1 mole = 6.023 $$ \times $$ 1023 number of molecules.

1 g mole of O2 = 32 g of O2

$$ \Rightarrow $$ 16 g of O2 = 0.5 g mole of O2

1 g mole of N2 = 28 g of N2

$$ \Rightarrow $$ 7 g of N2 = 0.25 g mole of N2

1 g mole of H2 = 2 g of H2

$$ \Rightarrow $$ 2 g of H2 = 1 g mole of H2

1 g of NO2 = 14 + 16 $$ \times $$ 2 = 46

$$ \Rightarrow $$ 16 g of NO2 = 0.35 mole NO2
4
MCQ (Single Correct Answer)

AIPMT 2001

Specific volume of cylinfrical virus particle is 6.02 $$ \times $$ 10$$-$$2 cc/g whose radius and length are 7 $$\mathop A\limits^ \circ $$ and 10 $$\mathop A\limits^ \circ $$ respectively. If NA = 6.02 $$ \times $$ 1023, find molecular weight of virus.
A
15.4 kg/mol
B
1.54 $$ \times $$ 104 kg/mol
C
3.08 $$ \times $$ 104 kg/mol
D
3.08 $$ \times $$ 103 kg/mol

Explanation

Specific volume ( vol. of 1 g) cylindrical virus particle

= 6.02 $$ \times $$ 10-2 cc/g

Radius of virus, r = 7 Å = 7 $$ \times $$ 10-8 cm

Volume of virus = $$\pi {r^2}l$$

= $${{22} \over 7} \times (7 \times {10^{ - 8}}) \times 10 \times {10^{ - 8}}$$

= 154 $$ \times $$ 10-23 cc

wt. of one virus particle = $${{Volume} \over {Specific\,\,Volume}}$$

$$ \Rightarrow $$ $${{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}}g$$

$$ \therefore $$ Molecular wt of virus = wt. of NA particle

= $${{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}} \times 6.02 \times 10^{-23} \,\,g/mol$$

= 15400 g/mol = 15.4 kg/mol

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