NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

AIPMT 2004

MCQ (Single Correct Answer)
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

AIPMT 2002

MCQ (Single Correct Answer)
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

AIPMT 2002

MCQ (Single Correct Answer)
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

AIPMT 2001

MCQ (Single Correct Answer)
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

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12