Q: Calculate the volume of 1 mole of gas exerting pressure of 1 atm at a temperature of 300 K.

PV=nRT ∴V=PnRT=1×0.0821×300=24.3 L

Q: 22.4 L of aq. {H2O] at STP contains how many moles?

1 mL=1 g 1 litre =1000 g=181000 moles

Q: How many atoms and molecules of sulphur are present in 64 g of S8 ?

1 mole of S8=NA molecules of S8=8×32 g ∴ number of molecules =8×3264=4NA ∴ number of atom of sulphur =8×4NA=2NA

Q: Calculate the number of molecules in (i) 34.2 g of cane sugar (C12H22O11) (ii) 1 L of water assuming density 1 g (iii) 1 drop of water having mass 0.05 g

(i) 34234.2×NA=10NA (ii) 1 mL=1 g ∴1000 mL=181000 gNA (iii) 180.05×NA=360NA

Q: Elemental analysis of an organic compound containing C,H,N and O and weighing 7.3 g was found to contain 3.6 g of carbon, 0.7 g of H and 1.4 g of nitrogen. Calculate Empirical formula.

H 0.710.7=0.7 C 3.6123.6=0.33 N1.4141.4=0.11 O1.61.61.6=0.11 Empirical formula =C3H7NO C Atomic weight weight H Atomic weight weight N Atomic weight weight O Atomic weight weight C123.6H10.7 N141.4O1.61.6=C3H7NO

Q: Calculate the Empirical formula of the following minerals that have the following composition: ZnSO4= 56.14H2O=43.86.

ZnSO456.14161.456.14=0.350.350.35=1 H2O43.861843.86=2.440.352.44=7 ZnSO4⋅7H2O

Q: Calculate Empirical formula of hydrocarbon if 1.4 g of this compound is on complete combustion. Gives 4.4 g of CO2 and 1.8 g of H2O.

CxHy+O2→xCO2+2yH2O∵1 mole of CO2 contains 1 mole of C∴44 gofCO2 contains 12 gram of C∴4.4 g of CO2 contain 1.2 g of C1 mole H2O contains 2 mole of H.18 gram H2O contains 2 mole =2 g∴1.8 gram H2O contains 2 mole

Q: Calculate the weight of KClO3 required and the weight of oxygen in product. If 1.49 g of KCl is obtained on heating, calculate the volume of oxygen at STP.

2KClO3→2KCl+3O2 Number of moles of KCl=74.51.49 2 moles of KCl produced from 2 moles of KClO3. ∴ mass of KClO3=n× molecular weight =74.51.49×122.5 and 2 moles of KClO3 is required for 3 moles of O2. number of moles of O2 obtained =23[n]=23×[74.51.49] mass of O2=23[74.51.49]×32 volume of oxygen at STP=23×74.51.49×22.4 L=0.672 L

Q: When 4 g of CaCO3 and sand mixture is treated with excess of HCl,0.88 g of CO2 is produced. Calculate % weight of CaCO3.

CaCO3+2HCl→CO2+H2O+CaCl2 Sand +HCl→ No reaction Number of moles of CO2=440.88=0.02 mole ∴ number of moles of CaCO3=0.02 moles ∴% mass of CaCO3= Weight of mixture 0.02×100×100 =40.02×100×100=50%

Question 1

Calculate the volume of 1 mole gas exerting pressure of 1 atm at a temperature of 273 K.

Accepted Answer

V=PnRT​=0.0821×273=22.4 L

Question 2

Calculate the volume of 1 mole of gas exerting pressure of 1 atm at a temperature of 300 K.

Accepted Answer

PV=nRT
∴V=PnRT​=1×0.0821×300=24.3 L

Question 3

44.8 L of oxygen contains how many moles of O2​ ?

Accepted Answer

Since T and P are not given, so number of mole cannot be calculated.

Question 4

22.4 L of aq. {H2​O] at STP contains how many moles?

Accepted Answer

1 mL=1 g
1 litre =1000 g=181000​ moles

Question 5

How many atoms and molecules of sulphur are present in 64 g of S8​ ?

Accepted Answer

1 mole of S8​=NA​ molecules of S8​=8×32 g
∴ number of molecules =8×3264​=4NA​​
∴ number of atom of sulphur =8×4NA​​=2NA​

Question 6

Calculate the number of molecules in
(i) 34.2 g of cane sugar (C12​H22​O11​)
(ii) 1 L of water assuming density 1 g
(iii) 1 drop of water having mass 0.05 g

Accepted Answer

(i) 34234.2​×NA​=10NA​​
(ii) 1 mL=1 g
∴1000 mL=181000 g​NA​
(iii) 180.05​×NA​=360NA​​

Question 7

Elemental analysis of an organic compound containing C,H,N and O and weighing 7.3 g was found to contain 3.6 g of carbon, 0.7 g of H and 1.4 g of nitrogen. Calculate Empirical formula.

Accepted Answer

H 0.710.7​=0.7
C 3.6123.6​=0.33
N1.4141.4​=0.11
O1.61.61.6​=0.11
Empirical formula =C3​H7​NO
C Atomic weight  weight ​H Atomic weight  weight ​
N Atomic weight  weight ​O Atomic weight  weight ​
C123.6​​H10.7​​ N141.4​​O1.61.6​​=C3​H7​NO

Question 8

Calculate the Empirical formula of the following minerals that have the following composition: ZnSO4​= 56.14H2​O=43.86.

Accepted Answer

ZnSO4​56.14161.456.14​=0.350.350.35​=1
H2​O43.861843.86​=2.440.352.44​=7
ZnSO4​⋅7H2​O

Question 9

Calculate Empirical formula of hydrocarbon if 1.4 g of this compound is on complete combustion. Gives 4.4 g of CO2​ and 1.8 g of H2​O.

Accepted Answer

Cx​Hy​+O2​→xCO2​+2y​H2​O∵1 mole of CO2​ contains 1 mole of C∴44 gofCO2​ contains 12 gram of C∴4.4 g of CO2​ contain 1.2 g of C1 mole H2​O contains 2 mole of H.18 gram H2​O contains 2 mole =2 g∴1.8 gram H2​O contains 2 mole

Question 10

Calculate the weight of KClO3​ required and the weight of oxygen in product. If 1.49 g of KCl is obtained on heating, calculate the volume of oxygen at STP.

Accepted Answer

2KClO3​→2KCl+3O2​
Number of moles of KCl=74.51.49​
2 moles of KCl produced from 2 moles of KClO3​.
∴ mass of KClO3​=n× molecular weight
=74.51.49​×122.5
and 2 moles of KClO3​ is required for 3 moles of O2​.
number of moles of O2​ obtained

=23​[n]=23​×[74.51.49​]
mass of O2​=23​[74.51.49​]×32

volume of oxygen at STP=23​×74.51.49​×22.4 L=0.672 L

Question 11

When 4 g of CaCO3​ and sand mixture is treated with excess of HCl,0.88 g of CO2​ is produced. Calculate % weight of CaCO3​.

Accepted Answer

CaCO3​+2HCl→CO2​+H2​O+CaCl2​
Sand +HCl→ No reaction
Number of moles of CO2​=440.88​=0.02 mole
∴ number of moles of CaCO3​=0.02 moles
∴% mass of CaCO3​= Weight of mixture 0.02×100​×100
=40.02×100​×100=50%

Question 12

A sample of 5 g of natural gas consisting of methane and ethylene was burnt in excess of O2​ yielding 14.5 g of CO2​. Calculate the weight % of ethylene.

Accepted Answer

CH4​=x g
C2​H4​=5−x g
Number of Moles of CH4​=x/16
Number of Moles of C2​H4​=5−x/28
CH4​+O2​→CO2​+2H2​O
C2​H4​+O2​→2CO2​+2H2​O
Number of Moles of CO2​=4414.5​
4414.5​=[285−x​×2]+[16x​×1]
∴% weight of ethylene =55−x​×100
∴ number of moles of O2​ obtained
=23​[n]=23​×[74.51.49​]
∴ mass of O2​=23​[74.51.49​]×32
∴ volume of oxygen at STP
=23​×74.51.49​×22.4 L=0.672 L

Question 13

In a rocket motor filled with butane, how many kg of liquid oxygen should be provided for each kg of butane for its complete combustion?

Accepted Answer

C4​H10​+213​O2​→4CO2​+5H2​O
[1×103 g]
∴ number of moles of C4​H10​=581000​
∴ number of moles of O2​=[581000​]×213​
∴ Mass of O2​=[581000​]×213​×32 gram =3.586Kg

Question 14

In a sample of NaCl and KCl weight of mixture is 20.75 g. This was dissolved in water and heated with excess of AgNO3​. If weight of white precipitate of AgNO3​ is 43.05 g, calculate the weight of each component.

Accepted Answer

Weight of NaCl=x g
∴ weight of KCl=20.75−x g
58.5x​= (a mole)
KClAgNO3​⟶H2​O​​AgCl
143.520.75−x​=(b mole )
a+b=143.543.05​=58.5x​+74.520.75−x​

Question 15

Calculate the mass of compound [K2​Zn3​[Fe(CN)6​]2​ when 2.76 g of K2​CO3​ undergoes series of reaction such that all carbons convert into this complex. Given Molar Mass of complex compound (C.C) =698.2 and MM of K2​CO3​=138

Accepted Answer

K2​CO3​→K2​Zn3​[Fe(CN)6​]2​
Applied POAC on carbon
1 mole of K2​CO3​=12× moles of complex
MM2​ of K2​CO3​ Weight of K2​CO3​​=12[ MM of C.C  Weight of C.C ​]
1382.76​=12[698.2 Weight ​]∴ Weight =1.163 gram

Question 16

If 20%Vv​ is given then convert into %Ww​ and %Vw​. If d is the density of solution.

Accepted Answer

100 mL of solution contains 20 mL of NaOH and 80 mL of H2​O.
%Ww​= Mass of solution  Mass of solution − Mass of solvent ​×100
=[100d100d−80×1​]×100=[d100d−80​]%
Vw​%=[100( Volume of solution )100d−80×1​]×100=[100d−80]%

Question 17

Given 20%Vw​ of aq. NaOH. Then find its molarity.

Accepted Answer

100 mL solution contains 20 gNaOH.
∴1000 mL solution contains 200 gNaOH
∴ number of moles =40200​=5 moles and 1000 mL=1 L
∴ molarity =15​=5M

Question 18

Convert 40%Ww​NaOH solution into molarity. If density of solution is d g/mL.

Accepted Answer

100 g solution contains 40 gNaOH
40 g of NaOH=1 mole
Volume of solution =d×1000100​=10d1​ L
[ Since V= density  Mass ​]
∵10d1​ L solution contains 1 mole of NaOH
∴1 L solution contains 1 mole of NaOH
=10d1​1​=10dM
∴ Molarity =10dM

Question 19

Calculate molality of 20% by mass NaOH solution.

Accepted Answer

20% by mass means 20 gNaOH present in 80 g of solvent.
∴ Molality m=80/100020/40​=4020​×801000​

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Solutions for Physical Chemistry I for IIT JEE Main and Advanced by Ranveer Singh

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