1

### 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
2

### AIPMT 2001

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
3

### AIPMT 2001

Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular weight of peroxide anhydrous enzyme is
A
1.568 $\times$ 104
B
1.568 $\times$ 103
C
15.68
D
2.136 $\times$ 104

## Explanation

${{100} \over {0.5}} \times 78.4 = 1.568 \times {10^4}$
4

### AIPMT 2001

Molarity of liquid HCl, if density of solution is 1.17 g/ce is
A
36.5
B
18.25
C
32.05
D
42.10

## Explanation

Molarity = ${{1.17 \times 1000} \over {36.5 \times 1}} = 32.05$