1

### AIPMT 2008

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g HCl ?
A
0.011
B
0.029
C
0.044
D
0.333

## Explanation

PbO + 2HCl $\to$ PbCl2 + H2O
n mol 2n mol n mol
$6.5 \over 224$ mol $3.2 \over 36.5$ mol
0.029 mol 0.087 mol

Formation of moles of lead(II) chloride depends upon the no. of moles of PbO which acts as a limiting reagent here. So no of moles of PbCl2 formed will be equal to the no of PbO (i.e 0.029)
2

### 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

### 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

### 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.