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Mole and its relation with mass, volume and number of particles

11 Science Chemistry Very Short Question
40 Questions

Showing 15 of 35 (Results for all questions)

1 (1)

Calculate mass in gram of two atom of carbon.

Ans: (i) 3.98 x 10-23 g carbon

2 (2)

How many numbers of moles of CO will be left when 2 x 1021 molecules are removed from 0.28g of CO?

Ans: 6.68 x 1023

3 (1)

Mass of 2 x 1021 number of atoms of an element is 0.4g. What is the mass of 0.5 mole of the element?

Ans: 60.22g

4 (1)

Mass of 3.011 x 1022 molecules a gaseous substance is 22g. Calculate the volume of 33g of the gas at NTP.

Ans: 1.68 L

5 (1)

How many number of hydrogen atoms are present in:

  1. 64 amu of methane
  2. 4.5g of water ?

Ans: (i) 9.69 x 1024 (ii) 3.011 x 1023

6 (1)

What mass of nitrogen will be required to produce 6.022 x 1024 molecules of ammonia by the following reaction?

N2 + 3H2 => 2NH3

Ans: 1.4 x 1048g

7 (1)

How many number of oxygen molecules are required to produce 220 mg of CO2according to the reaction?

C + O2 => CO2

Ans: 3.011 x 1021 O2

8 (1)

Calculate the number of hydrogen and oxygen molecules in a solution prepared by mixing 45g of glucose in 36g of water.

Ans: H2 = 2.1077 x 1024 , O2 = 1.05385 x 1024

9 (1)

How many grams of sulphur and oxygen are needed to produce 6.022 x 1024 molecules of SO2 according to the reaction?

S + O2 => SO2

Ans: 320g sulphur and 320g oxygen

10 (2)

Calculate mass in gram of three molecules of hydrogen.

Ans: 9.96 x 10-24 g

11 (1)

Why one has higher mass and why?0.5 mole of CO2 or 16g of SO2.2g of hydrogen or 6.023 x 1021 number of oxygen molecules.

Ans: (a) 0.5 mol of CO2 (b) 2g of H2

12 (1)

Calculate the number of molecules of hydrogen and carbon present in 4g of methane.

Ans: H2 = 3.011 x 1023 ; C = 1.5 x 1023

13 (1)

Calculate the mass of 4 atom of carbon

Ans: 7.97 x 10-23

14 (1)

The cost of per mole of sugar (C12H22O11) is Rs. 20. How much a pocket of sugar containing 2 kg would cost?

Ans: Rs. 116.95

15 (1)

Calculate the mass of carbon monoxide having the same number of oxygen atoms as are present in 88g of carbon dioxide.

Ans: 56g CO

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