Question Pool Gas Laws & Mole Concept

Gas Laws & Mole Concept - SAMAGRA Question Pool & Answers | Class 10 English Medium

Kerala Syllabus SAMAGRA SCERT SAMAGRA Question Pool for Class 10 English Medium Chemistry Gas Laws & Mole Concept

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Qn 1.

The volume of a fixed mass of gas at 300K is 10L.What will be the volume of the gas,if the temperature is doubled without changing the pressure.

  • Answer)

    Volume and temperature are directly proportional.Therefore volume changes to 20L / Volume doubled

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Qn 2.

The relation showing the volume and temperature of fixed mass of gas at constant pressure is tabulated below.

Volume V(L)

Temperature T(K)

V / T

600

300

2

800

.......(a).....

2

......(b)........

450

2

 

i) Find out the values of a and b.

ii)State the gas law associated with this.

iii) Write down any one instance from daily life related with this law.

  • Answer)

    i) a = 400, b = 900

    ii) At constant pressure,the volume of a definite mass of a gas is directly proportional to the temperature in

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Qn 3.

a) What happens to the size of a gas bubble rising from the bottom of a water body?why?

b)Which is the gas law associated with this?

  • Answer)

    a)size increases

    As the bubbles move from bottom to top in a water body,pressure decreases and correspondingly the volume increases.

    b)Boyle's

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Qn 4.

The volume of a fixed mass of gas at 2 atm pressure is 20L.What will be its volume if the pressure is increased 4 times without changing the temperature.

  • Answer)

    PV = a constant

    2 x20 = 40

    8 x X = 40

    X = 40 / 8 =5

    Volume changes to 5

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Qn 5.

The data of an experiment conducted on a fixed mass of gas at constant temperature are given

Pressure P(atm)

Volume V(L)

PV

1

10

.....(a)....

2

......(b)...........

10

.....(c).........

2.5

10

 

i)Complete the table and find out the speciality of PV.

ii)What is the relation between pressure and volume?

iii) Which gas law can be proved by this experiment?

  • Answer)

    i) a = 10,b = 5L, c = 4 atm, PV ia a constant

    ii)Volume and pressure are inversely propotional.

    iii)Boyle's

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Qn 6.

What happens to the following when the temperature of a gas in a closed container is increased ?

a) Kinetic energy

b)Pressure

  • Answer)

    a) Kinetic energy increases

    b)Pressure

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Qn 7.

When a gas contained in a 2L cylinder is completely transferred to a 4L cylinder,the volume of the gas will be ...........

  • Answer)

    4L

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Qn 8.

Select the statements suitable to gases from those given below.

a) Intermolecular distance is very low.

b)The volume of gas depends on the volume of the container in which it is occupied.

c)The energy of gaseous molecules is very high.

d)The attractive force between gaseous molecules is very high.

  • Answer)

    b)The volume of gas depends on the volume of the container in which it is occupied

    c / The energy of gaseous molecules is very

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Qn 9.

a) How many moles are there in 140g Nitrogen?.

b) How many atoms are there in 140g Nitrogen?

(Atomic mass : N- 14 )

  • Answer)

    (a) 5

    (b) 10

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Qn 10.

Find out the molecular mass of the following compounds

(Atomic Mass : Ca - 40 , N- 14 , C - 12 , O -16 , H- 1)

a) Ca(NO3)2           b) C12H22O11

  • Answer)

    a = 164, b = 342

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Qn 11.

N+ 3 H2→ 2NH3

The ratio of reactants and products in the above reaction is 1:3:2 .Complete the table related with this reaction.

 

Chemical reaction

Reactants

Products

N2

H2

NH3

Moles

(a)

6

4

Molecules

4 x 6.022 x1023

(b)

8 x 6.022 x1023

Volume at

STP

(c)

69.2 L

44.8 L

Mass

140 g

30 g

(d)

  • Answer)

    a) 2

    b) 12 x6.022 x1023

    c) 22.4 L

    d) 170

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Qn 12.

NaOH + HCl → NaCl + H2O

a) How many moles of NaOH is needed to completely react with 1 mole of HCl ?

b) How many grams of HCl is required to completely neutralise 160g NaOH ?

  • Answer)

    a) 1

    b) 146 g

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Qn 13.

Analyse the following equation and answer the questions

2Na + Cl2→ 2NaCl

a) What is the ratio of reactant molecules and product molecules?

b)How many moles of NaCl will be obtained on reaction of 10 moles of chlorine ?

c) Find the mass of sodium required to get so much amount of NaCl .

  • Answer)

    a) 2:1:2

    b)20mole

    c) 20x 23 =

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Qn 14.

 i) (i)Find a,b and c

ii) How many grams of H2O is required to get 5 x 6.022 x 1023 molecules ?

  • Answer)

    i)

    a) 10 x 6.022 x 1023

    b) 224 L

    c) 10 GMM

    ii)

    90 g

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Qn 15.

Which of the following have the same number of moles ?

[4 GMM H2 , 88 g CO2 , 89.6 L O2 , 4 g He]

  • Answer)

    4 GMM H2 , 89.6 L

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Qn 16.

Which one contains 2 x 6.022 x1023 Molecules ?

(28 g N2 , 2 g H2 , 32 g O2 , 44.8 L CO2)

  • Answer)

    44.8 L CO2

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Qn 17.

Which one is used as the basis of atomic mass now a days?

(H-1 , C-12 , C-14 , O – 16)

  • Answer)

    C-12

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Qn 18.

4 x 6.022 x 10 23 Chlorine molecules at STP are taken. Answer the following

questions(Atomic mass : Chlorine = 35.5)

a) What is its volume at STP ?

b) What is the mass of this compound?

c) H2 + Cl2 → 2HCl

How many molecules of hydrogen are required to completely react with 4 x6.022x1023 molecules of chlorine ?

  • Answer)

    a) 89.6 L

    b) 284 g

    c) 4 x

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Qn 19.

Volume of 2 x 6.022x1023 molecules of a gas at STP is _____

  • Answer)

    2 x 22.4L = 44.8 L

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Qn 20.

Mass of ¼ x 6.022x1023 Oxygen molecule is _____________ .

(Hint : Oxygen- Molecular mass = 32)

  • Answer)

    8 g

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Qn 21.

Complete the table.

Substance

Volume at STP

Number of moles

Mass(g)

CO2

44.8 L

2

88

CH4

(a)

(b)

4 g

NH3

11.2 L

(c)

(d)

(Hint : MM : CO= 18 , CH= 16 , NH3 = 17 )

  • Answer)

    a) 1/4 x 22.4 = 5.6 L

    b) ¼ or 0.25

    c) ½

    d) 8.5

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Qn 22.

N2 + 3 H2 → 2 NH3

Number of moles of hydrogen required to completely react with 2moles of nitrogen is _______

  • Answer)

    6 mole hydrogen

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Qn 23.

360 g glucose [C6H12O6] is given.

a) How many molecules are there in the sample ?

b) What is the total number of atoms in the sample? (Hints: Molecular mass C6H12O6 = 180)

  • Answer)

    a) GMM of C6H12O= 180 g

    Number of moles in 360g glucose = 360g / 180 g = 2

    Number of molecules = 2 x

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Qn 24.

Which of the samples given below contains 1mole Oxygen atoms ?

(Atomic mass O = 16 )

a. 16 g Oxygen .

b. 8g Oxygen.

c . 32 g Oxygen.

d . 22.4 L oxygen at STP

  • Answer)

    a. 16 g Oxygen.

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Qn 25.

Some samples are given

(P) 22.4 L NH3 (Q) 22 g CO2 (R) 64 g SO2 (S) 117 g NaCl

(GMM : NH3 = 17 g , CO= 44 g (c) SO2 = 64 g (d) NaCl = 58.5 g)

a) Which among the above are having the same moles?

b) How many molecules are there in sample Q?

c) How many grams of NHare needed to get the same number of molecules in sample S ?

  • Answer)

    a) P, R

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Qn 26.

Which among the following samples have the same number of molecules.

a) 88 g CO2 b) 54 g H2O c) 4 g Hd) 17 g NH3

(Atomic mass : C = 12 , O = 16 , H = 1 ,N =14 )

  • Answer)

    a, c

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Qn 27.

The equation showing the reaction of Zinc with hydrochloric acid is given.

Zn + 2HCl → ZnCl2 + H2

a) How many molecules of ZnCl2 will formed on complete reaction of 65g Zn with

HCl?

b) What will be the volume of Hformed at STP when 6.5g Zn reacts with HCl.

(Hint:Atomic mass : Zn = 65 , Cl = 35.5 , H= 1 )

  • Answer)

    a) 6.022x1023 (1 mole)

    b) 0.1 x 22.4 litre = 2.24

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Qn 28.

2Mg + O2 → 2MgO

The equation showing the burning of Magnesium is given. suppose 120g of Mg is burned.

a) How many atoms are there in 120g Mg ?

b) How much will be the volume of oxygen at STP to burn this much Mg?

c ) What will be the mass of Magnesium Oxide formed ?

(Hint : Atomic mass : O = 16, Mg = 24 )

  • Answer)

    a) (120/24) x 6.022x1023 = 5 x 6.022x1023

    b) 5/2 x 6.022x1023

    c) 5 x (24+16) = 5 x 40 g =

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Qn 29.

Match the following.

A

B

C

10 g H2

3 x 6.022x1023

2 mol atoms

54 g H2O

2 GAM

112 L at STP

32 g O2

5 x 6.022x1023

3 GMM

  • Answer)

     



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    Qn 30.

    H2 + Cl2  →   2HCl

    The above experiment is carried out by using 10g H2 and 142g Cl2.

    a ) How many molecules are there in 142g of Cl2.

    b ) what is the volume of each of the above gaes at STP?

    c ) How many molecules of HCl will be formed in the reaction ?

    (Hint : Atomic mass : H = 1 , Cl = 35.5)

    • Answer)

      a) 2 x 6.022x1023

      b) H- 5 x 22.4 L = 112L

      Cl2 - 2 x 22.4L = 44.8 L

      c) 4 x

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    Qn 31.

    Choose the correct statements from those given below

    a) The volume of a mole of gas at 300K and 1atm is 22.4 L .

    b) 1GMM of any substance contains 6.022x1023molecules.

    c) The mass of 6.022x1023 O2 molecules is 16g .

    d ) The mass of 22.4L of oxygen at 273K and 1atm pressure is 32 g

    • Answer)

      statements b,d .

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    Qn 32.

    Arrange the following samples in the increasing order of their mass.

    a) 5 GMM CO2

    b) 10 GMM Oxygen

    c) 2 mol H2O

    d) 3 mol N2

    (Hint: Molecular mass- CO2 =44,O2=32,H2O=18, N2=28)

    • Answer)

      a=220g,b=320g,c=36g,d=84g

      c < d < a <

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    Qn 33.

    Arrange the following samples in the ascending order of number of moles.

    a) 90 g H2O

    b) 48 g CH4

    c) 100 g CaCO3

    d) 96 g SO2

    (Hint:Molecular mass- H2O =18,CH= 16,CaCO3 =100,SO=64)

    • Answer)

      a = 5,b=3,c=1 d=1.5

      c < d < b <

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    Qn 34.

    Complete the table. (Hint : atomic mass : He = 4 , N=14 , O =16 , P = 31 )

    Substance

    Atomic mass

    Amount taken(g)

    No.of molecules

    No. of atoms

    He

    4

    10

    (a)

    (b)

    N2

    14

    (c)

    6.022x1023

    (d)

    Cl2

    35.5

    (e)

    (f)

    10x 6.022x1023

    O2

    (g)

    80

    (h)

    5x 6.022x1023

    • Answer)

      a = 2.5 x 6.022x1023

      b= 2.5 x 6.022x1023 

      c = 28g

      d= 2 x 6.022x1023 

      e = 355 g

      f= 5

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    Qn 35.

    Arrange the following samples in the increasing order of number of atoms.

    (hint : atomic mass : H = 1 C = 12 O =16 Ca = 40 )

    a ) 10 g Hydrogen b ) 100 g Calcium c ) 64g Oxygen d ) 36g Carbon

    • Answer)

      a) 10 GAM b) 2.5 GAM c) 4 GAM d) 3GAM

      b < d < c <

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    Qn 36.

    1mL of oxygen at constant temperature and pressure contains  x molecules.

    write answer related to the following gases at same temperature and pressure.

    a) Number of molecules in 1mL hydrogen?

    b)Number of molecules in 5mL nitrogen ?

    c)Volume of 3x molecules of CO2 ?

    • Answer)

      a = x, b = 5x , c = 3mL

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    Qn 37.

    Choose the correct statements from those given below .

    (Hint : Atomic mass : C - 12 , O - 16 )

    a) 6.022 x 1023 molecules are there in 22 g CO2.

    b) 1 GMM of CO2 is 22 g .

    c) Volume of 22 g CO2 at STP is 11.2 L.

    d) 22 g of CO2 contains3 x  ½ x 6.022 x 1023 atoms.

    • Answer)

      c,d

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    Qn 38.

    Pick the odd one out ?

    64 g SO2 , 2 x 6.022 x 1023 H2 molecules , 64 g O2 , 44.8 L CO2 at STP

    (Atomic mass : S - 32 , O -16)

    • Answer)

      64 g SO2

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    Qn 39.

    Find a,b,c .

    (Hint: MM- CH4 =16)

    • Answer)

      a) ½ GMM

      b) ½ x 5 x  6.022x1023

      c) 8

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    Qn 40.

    The mathematical representation of some gas laws are given. Identify the law related to each one.

    a) V∝T

    b)V ∝ 1/p

    c)V∝n

    • Answer)

      a)Charles' law

      b) Boyle's law

      c)Avogadro's

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    Qn 41.

    Find out the gas law related with each of the followig instances.

    a)The size of the balloon increases as it is inflated.

    b)An inflated balloon kept in  direct sunlight is found to burst.

    c)Gases can be marketed in cylinders.

    • Answer)

      a )Avogadro's Law

      b) Charles' law

      c) Boyle's

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    Qn 42.

    An inflated balloon contains X air molecules.After some time the volume of the balloon is found to be the half at the same temperature and pressure when a few  air molecules are expelled out.

    a)How many molecules will be there in the balloon now? 

    b) Which is the gas law associated with this?

    • Answer)

      a = X /2,

      b -Avogadro's Law

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    Qn 43.

    The mass of 5 GAM X is 80g . [Symbol is not real]

    a ) What is the atomic mass of this element ?

    b ) How many atoms are there in 80g X?

    c )How many grams of helium are to be taken to get as many molecules as there in  X?

    (Atomic mass : He = 4 )

    • Answer)

      a) 16

      b) 5 x 6.022x1023

      c) 20

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