1
MCQ (Single Correct Answer)

AIPMT 2007

An element, X has the following isotopic composition:
200X : 90%  199X : 8.0%  202X : 2.0%

The weighted average atomic mass of the naturally occuring element X is closest to
A
201 amu
B
202 amu
C
199 amu
D
200 amu

Explanation

Average isotope mass of X

= $${{200 \times 90 + 199 \times 8 + 202 \times 2} \over {90 + 8 + 2}}$$

= $${{18000 + 1592 + 404} \over {100}}$$

= 199.96 a.m.u $$ \approx $$ 200 a.m.u
2
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
3
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.
4
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

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