Malachite has the formula Cu₂CO₃(OH)₂. What percentage by mass of malachite is copper?

25%
50.9%
57.5%
63.5%

Solution

Percentage by mass of copper in malachite
= \(\frac{2 \times 63.5}{221}\) =57.5%

In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product.What will be the composition of gaseous mixture under the aforesaid condition in the end?

20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen

Solution

Percentage of Se in peroxidase anhydrase enzyme is 0.5% by weight (at. wt. of Se = 78.4) then minimum molecular weight of peroxidase anhydrase enzyme is

1.568 × 10³
15.68
2.136 × 10⁴
1.568 × 10⁴

Solution

0.5 g of Se = 100 g enzyme

78.4 g of Se = \(\frac{11}{0.5}\) × 78.4 = 1.568 × 10⁴ g enzyme

Specific volume of cylindrical virus particle is 6.02 × 10^–2cc/gm.whose radius and length 7 Å & 10 Å respectively.If NA= 6.02 × 1023, find molecular weight of virus

3.08× 10³kg/mol
3.08 × 10⁴kg/mol
1.54 × 10⁴kg/mol
15.4 kg/mol

Solution

Haemoglobin contains 0.334% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. wt.of Fe is 56) present in one molecule of haemoglobin are

Solution

Mass of iron in 100 g haemoglobin = 0.334 g
∴ In 67200 g haemoglobin, mass of iron
=\(\frac{67200 \times 0.334}{100}\) = 672 × 0.33g

∴ The number of Fe atoms in one molecule of haemoglobin

= \(\frac{672 \times 0.334}{56}\) = 4

The density of 0.5 M glucose solution is 1.0900 g ml^–1. The molality of the solution is

0.1000
0.2000
0.2500
0.5000

Solution

Mass of 1 L (= 1000 ml)solution =1000 × 1.090 = 1090g
Mass of glucose in 1L = 0.5 × 180 = 90 g.
Mass of water = 1090.0 g – 90.0 g= 1000 g
Hence, molality = \(\frac{0.5 \times 1000}{1000}\) = 0.5

Percent by mass of a solute (molar mass = 28 g) in its aqueous solution is 28. Calculate the mole fraction (X) and molality (m)of the solute in the solution.

X = 0.2, m= 10
X= 0.2, m = 125/9
X = 0.8, m = 125/9
X = 0.8, m = 10

Solution

Ratio of C_p and C_v of a gas ‘X’ is 1.4. The number of atoms of the gas ‘X’ present in 11.2 litres of it at NTP will be

6.02 ×10²³
1.2 × 10²³
3.01× 10²³
2.01 × 10²³

Solution

Cp/ Cv= 1.4 shows that the gas is diatomic.
22.4 litre at NTP = 6.02 ×10²³ molecules
11.2 L at NTP = 3.01 × 10²³ molecules
= 3.01 × 10²³ × 2 atoms = 6.02 × 10²³ atoms

A metal oxide has the formula Z₂O₃. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

27.9
159.6
79.8
55.8

Solution

The specific heat of a metal is 0.16 cal g^–1. The equivalent mass of the metal is 20.04, the correct atomic mass of the metal is :

40
20.04
40.08 g
80.16 g

Solution

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