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1

### 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
2

### 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

$${{100} \over {0.5}} \times 78.4 = 1.568 \times {10^4}$$
3

### 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 = $${{1.17 \times 1000} \over {36.5 \times 1}} = 32.05$$
4

### AIPMT 2000

Oxidation numbers of A, B, C are + 2, +5 and $$-$$2 respectively. Possible formula of compound is
A
A2(BC2)2
B
A3(BC4)2
C
A2(BC3)2
D
A3(B2C)2

## Explanation

In A3(BC4)2, (+2) $$\times$$ 3 + 2[+5+4(-2)]

$$\Rightarrow$$ +6 + 10 -16 = 0

$$\therefore$$ Oxidation numbers of A, B, C are + 2, +5 and $$-$$2 respectively.

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