Kerala SCERT Class 10 Chemistry Chapter 2 Gas Laws And Mole Concept Question Answer Solution Here. Kerala Board Class 10 Students can find Here 1st Chapter Gas Laws And Mole Concept Notes provide by our Teacher. Important Question Answer from Chapter 2 Gas Laws And Mole Concept.
- Board- Kerala Board.
- Class – 10.
- Subject – Chemistry Part 1
- Chapter – 2
- Chapter Name – Gas Laws And Mole Concept.
- Topic – Question Answer Solution.
Q.1) MCQ
1) Energy of gas molecules is …………
a) low
b) high
c) moderate
d) None of the above
2) Attractive force between gaseous molecules is ……….
a) low
b) high
c) moderate
d) No attractive force
3) Volume of gas is ……….
a) volume of a container
b) volume of a one gas molecule
c) None of the above
4) Temperature is average ……………..energy of molecule.
a) Potential Energy
b) Kinetic Energy
c) Total Energy
d) Total Force
5) Which of the following equation is correct.
a) PV = constant
b)= constant
c) VT = constant
d) None of above.
6) In Charle’s law, which relationship is given.
a) Volume and pressure
b) Volume and temperature
c) Pressure and temperature
d) None of the above
7) According to Avogadro’s law, volume of a gas is directly proportional to…………
a) Pressure
b) Temperature
c) Number of molecules
d) None of the above
8) Atomic mass of carbon is……….
a) 11
b) 12
c) 13
d) 14
9) 1 mole of hydrogen contains …………
a) 1 gm
b) 6.022 x 1023 atoms
c) 100 gm
d) 10 gm
10) Standard temperature is……….
a) 272 K
b) 273 K
c) 274 K
d) 275 K
Q.2) Very Short Answer
1) Define volume of gas
2) Define pressure and give its equation.
3) Define temperature.
4) State Boyle’s law.
5) State Charle’s law.
6) State Avogadro’s law.
7) State definition of Avogadro’s number.
8) What is one mole atom.
9) Define Gram molecular mass.
10) What is one mole molecule.
Q.3) Short Answer Question.
1) Calculate molecular mass of glucose.
(C6H12O6) (Atomic mass C=12, H=1, O=16)
2) Complete the table
| Element | Molecular mass | Mass in gms | No. of molecules |
| Hydrogen (H2) | ….. | ….. | ….. |
| Oxygen (O2) | ….. | ….. | ….. |
| Nitrogen (N2) | ….. | ….. | ….. |
3) How many molecules of H2O present in 1 GMM. And give the Molecular mass.
4) Complete the table.
| Volume V | Temperature T (inkelvine) | V T |
| 546 mL | 273 K | ….. |
| 600 mL | 300 K | ….. |
| 660 mL | 330 K | ….. |
5) The molecular mass of ammonia (NH3) is 17.
a) How much is the GMM of ammonia.
b) find out no. of molecules present in 170 g of ammonia.
c) Calculate number of ammonia molecules present in above sample of ammonia.
Q.4) Long Answer Question
1) a) Calculate mass of 112 L CO2 gas at STP (molecular mass – 44)
b) How many CO2 molecules present ?
2) Calculate volume of 170 g of Ammonia at STP (molecular mass = 17)
3) Complete the table
| Element Compound | Molecular mass | Mass in grams | GMM | No. of molecules |
| 2 moles (H2O) | ….. | ….. | ….. | ….. |
| 1 mole (NH3) | ….. | ….. | ….. | ….. |
| 2 mol. (O2) | ….. | ….. | ….. | ….. |
| 3 mol. (H2) | ….. | ….. | ….. | ….. |
| 1 mol. (N2) | ….. | ….. | ….. | ….. |
4) The molecular mass of NH3 and O2 is 17 and 32 respectively.
a) How is the GMM for both.
b) find out number of moles present in 170 g of NH3 and 64 g Oxygen respectively.
ANSWER SHEET
GAS LAWS AND MOLE CONCEPT
Q.1) MCQ
1) = b)
2) = d)
3) = a)
4) = b)
5) = a)
6) = b)
7) = c)
8) = b)
9) = b)
10) = b)
Q.2) Very Short Answer
1) Volume of a gas is the volume of a container which it occupies.
2) Force exerted per unit area is called as pressure.
Pressure = Total force exerted on the surface
Surface area
3) Temperature is the average kinetic Energy of molecules in a substance
4) Boyle’s Law: – At constant temperature, volume of a definite mass of A gas is inversely proportional to its pressure If P is pressure and V is
Volume, the P x V is constant.
5) At constant pressure, the volume of a definite mass of a gas is directly Proportional to temperature in kelvin.
V/T = constant
6) At constant temperature and pressure, the volume of a gas is directly Proportional to the number of molecules.
7) One gram atomic mass of any element contains 6.022 x 1023atoms.
This number is known as avogadro’s number. This is indicated as NA.
8) One mole atom is equal to 6.022 x 1023 atoms.
e.g.: – 12 g c = 1 GAM C = 6.022 x 1023 atoms = 1 mole of C.
9) The amount of a substance in grams is equal to its molecular mass is Called Gram Molecular mass.
10) One mole of molecules means 6.022 x 1023molecules.
1GMM = 1 mole = 6.022 x 1023 molecules.
Q.3) Short Answer Question.
1) Molecular mass of Glucose
C6H12O6 = C x 6 + H x 12 + 0 x 6
= 12 x 6 + 1 x 12 + 96
= 180 g
2) Complete the table.
| Elements | Molecular mass | Mass in gm | No. of molecules |
| Hydrogen (H2) | 2 | 2 g | 6.022 x 1023 |
| Oxygen (O2) | 32 | 32 g | 6.022 x 1023 |
| Nitrogen (N2) | 28 | 28 g | 6.022 x 1023 |
3) molecular mass of a substance is One Gram Molecular Mass.
Molecular mass of H2O = H x 2 + O x 1
= 1 x 2 + 16 x 1
= 2 + 16
= 18 g
1 GMM = 18 gm of H2O = 6.022 x 1023 H2O molecules.
4) Complete the table
| Volume V | Temperature T (in Kelvine) | V T |
| 546 mL | 273 k | 2 |
| 600 mL | 300 k | 2 |
5) a) GMM i.e., Gram Molecular Mass of ammonia is 17 gm.
b) 17 gm NH3 = 6.022 x 1023 molecules
170 gm NH3 =?
X = 6.022 x 1023 x 170
17
= 6.022 x 1024
c) 170 gm of NH3 = 170 x molecular mass of NH3
= 170 x 17
= 2890 g
Q.4) Long Answer Question.
1) a) No of moles = (Volume at STP)/22.4-
= 112/22.4
= 5 mol.
Mass of CO2 = mol x GMM
= 5 x 44 g = 220 g
b) No of molecules in 112 L of CO2 gas
= 112/22.4 x 6.022 x 1023
= 5 x 6.022 x 1023
= 3.011 x 1024
2) 1 GMM = 17 g NH3 = 22.4 L
170 g NH3 = x
= (170 x 22.4)/17
Volume = 224 L
3) Complete the table
| Element Compound | Molecular mass | Mass in grams | GMM | No. of molecules |
| 2 moles (H2O) | 36 | 36 g | 2 GMM | 2 x 6.022 x 1023 |
| 1 mole (NH3) | 17 | 17 g | 1 GMM | 6.022 x 1023 |
| 2 mol. (O2) | 64 | 64 g | 2 GMM | 2 x 6.022 x 1023 |
| 3 mol. (H2) | 6 | 6 g | 3 GMM | 3 x 6.022 x 1023 |
| 1 mol. (N2) | 28 | 28 | 1 GMM | 6.022 x 1023 |
4) a) NH3 = molecular mass of NH3 and O2 is 17 and 32 respectively,
1 GMM = 17 gm of NH3
O2 = molecular mass = 32
1 GMM = 32 g of O2
b) Number of molecules present in 170 g NH3
No. of molecules = 170/17
= 10 molecules of NH3
No. of molecules present in 64 g O2
No. of molecules = 64/32
= 2 molecules of O2
