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Some Basic Concepts of Chemistry · Question 43

AIPMT 2004MCQEasy📈 Topic in 50% papers+4−1
1

The maximum number of molecules is present in

  1. 15 L of H2 gas at STP
  2. 5 L of N2 gas at STP
  3. 0.5 g of H2 gas
  4. 10 g of O2 gas

Answer

(1) 15 L of H2 gas at STP

Solution

Number of molecules ∝ number of moles. At STP, 1 mol of gas occupies 22.4 L.

(1) n = 1522.4 = 0.67 mol

(2) n = 522.4 = 0.22 mol

(3) n = 0.52 = 0.25 mol

(4) n = 1032 = 0.31 mol

∴ 15 L of H2 at STP has the maximum number of moles and hence of molecules.

Answer: (1)

AIPMT 2004

Keep practising — Some Basic Concepts of Chemistry

2

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 ?

  1. 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
  2. 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
  3. 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
  4. 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
Out of syllabusAIPMT 2003
3

Which has maximum molecules?

  1. 7 g N2
  2. 2 g H2
  3. 16 g NO2
  4. 16 g O2
AIPMT 2002
4

The percentage of C, H and N in an organic compound are 40%, 13.3% and 46.7% respectively then empirical formula is

  1. C3H13N3
  2. CH2N
  3. CH4N
  4. CH6N
AIPMT 2002
5

2.5 litre of 1 M NaOH solution is mixed with another 3 litre of 0.5 M NaOH solution. Then find out molarity of resultant solution.

  1. 0.80 M
  2. 1.0 M
  3. 0.73 M
  4. 0.50 M
AIPMT 2002
6

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

  1. 1.568 104
  2. 1.568 103
  3. 15.68
  4. 2.136 104
AIPMT 2001
7

Molarity of liquid HCl, if density of solution is 1.17 g/cc, is

  1. 36.5
  2. 18.25
  3. 32.05
  4. 42.10
AIPMT 2001
8

Specific volume of cylindrical virus particle is 6.02 10-2 cc/g whose radius and length are 7 A and 10 A respectively. If NA = 6.02 1023, find molecular weight of virus.

  1. 15.4 kg/mol
  2. 1.54 104 kg/mol
  3. 3.08 104 kg/mol
  4. 3.08 103 kg/mol
AIPMT 2001
9

Volume of CO2 obtained by the complete decomposition of 9.85 g of BaCO3 is

  1. 2.24 L
  2. 1.118 L
  3. 0.84 L
  4. 0.56 L
AIPMT 2000

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