1

### 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

### 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

### 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

### 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