1
MCQ (Single Correct Answer)

AIPMT 2008

What volume of oxygen gas (O2) measuread at 0oC and 1 atm, is needed to burn completely 1 L of propane gas (C3H8) measured under the same conditions ?
A
5 L
B
10 L
C
7 L
D
6 L

Explanation

C3H8 + 5O2 $$ \to $$ 3CO2 + 4H2O (balanced equation)
1 vol. 5 vol. 3 vol. 4 vol.

According to the above equation 1 vol. or 1 litre of propane requires to 5 vol. or 5 litre of O2 to burn completely.
2
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
3
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
4
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.

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