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12th Grade > Chemistry

MOLE CONCEPT MCQs

Total Questions : 30 | Page 1 of 3 pages
Question 1. The volume of water that must be added to a mixture of 250 ml of 0.6 M HCl and 750 ml of 0.2 M HCl to obtain 0.25 M solution of HCl is
  1.    750 ml
  2.    100 ml
  3.    200 ml
  4.    300 ml
 Discuss Question
Answer: Option C. -> 200 ml
:
C
We can see that the number of moles of
HCl should remain constant, as per the law of conservation of mass.
Since we have molarity, we can calculate number of moles using,
No. of moles = Molarity × volume of solution
Initial Number of moles = 250 × 0.6 + 0.2 × 750
Final number of moles = 0.25 (250 + 750 + Vrequired) Equating, we get
Vrequired=0.6×250+0.2×7500.25[1000]=200ml
Question 2. How to determine mole fraction of solvent when moles of solute and solvent are given?
  1.    Moles of soluteMoles of solvent
  2.    Moles of soluteMoles of solution
  3.    Moles of soluteMoles of solute+ Moles of solvent
  4.    Moles of solventMoles of solute+Moles of solvent
 Discuss Question
Answer: Option D. -> Moles of solventMoles of solute+Moles of solvent
:
D
Mole fraction (X) is just the ratio of the number of moles of the required quantity to the sum of number of moles of all other quantities. We usually only have a solute and a solvent.
Xsolvent=MolesofsolventMoles of solute + Moles of solvent
Question 3. Calculate Molarity of a solution having 3.42 gm sucrose in 500 mL - solution. 
  1.    0.2 M
  2.    0.02 M
  3.    2 M
  4.    0.342 M
 Discuss Question
Answer: Option B. -> 0.02 M
:
B
We know that,
Molarity=No. of moles of soluteNo of litres of solution
We know the given weight of sucrose from which the number of moles of sucrose can be calculated. Therefore No. of moles =GivenweightMolecularweight(solute is sucrose)
=3.42342=0.01
When 0.01 moles of sucrose is added to 500ml
Molarity=0.01×1500×103(1mL=103L)
Molarity = 0.02 M
Question 4. What is the n-factor / valency factor of H2CO3 ?
  1.    1
  2.    2
  3.    Either a or b
  4.    Neither a nor b
 Discuss Question
Answer: Option C. -> Either a or b
:
C
In this case of H2CO3 valency factor changes depending upon the reaction as the number of H+ ions replaced might vary depending on the extent of the reaction.
H2CO3 H+ +HCO3 (n-factor =1)
H2CO3 2H+ +CO32 (n-factor = 2)
Question 5. How much sugar will be required if each person on earth is given 100 molecules of sugar, if the population of earth is taken to be 3 × 1010?
  1.    170.43 × 10−10 g
  2.    170.43 × 10−11g
  3.    340.86 × 10−11 g
  4.    340.86 × 10−10 g
 Discuss Question
Answer: Option B. -> 170.43 × 10−11g
:
B
No. of molecules required = 3×1010×102=3×1012
6×1023 molecules weight 342 g
3×1012 molecules weight?
170.43×1011g
Question 6. H3PO4 is a tribasic acid and one of its salt is NaH2PO4. What volume of 1M NaOH solution should be added to 12 g ofNaH2PO4 to convert it intoNa3PO4
  1.    100 ml
  2.    200 ml
  3.    80 ml
  4.    300 ml
 Discuss Question
Answer: Option B. -> 200 ml
:
B
NaH2PO4 basicity = 2
No of moles ofNaH2PO4 =12120
ENaOH=ENaH2PO4
V×1M =12120×1000×2
V =200ml
Question 7. How many carbon atoms are present in 0.35 mol of C6H12O6 :
  1.    6.023×1023 carbon atoms
  2.    1.26×1023 carbon atoms
  3.    1.26×1024 carbon atoms
  4.    6.023×1024 carbon atoms
 Discuss Question
Answer: Option B. -> 1.26×1023 carbon atoms
:
B
1moleofC6H12O6has=6NAatomsofC0.35moleofC6H12O6has=6×0.35NAatomsofC=2.1NAatoms
=2.1×6.022×1023=1.26×1024 carbon atoms
Question 8. In the species
i) 4.0g of NaOH          ii) 4.8g of SO2
iii) 4.0 g of CO2         iv) 2.8 g of CaO
decreasing order of the weight of oxygen is
  1.    iii, ii, i & iv
  2.    i, ii, iv & iii
  3.    ii, iii, iv & i
  4.    ii, iii, i & iv 
 Discuss Question
Answer: Option A. -> iii, ii, i & iv
:
A
4.0g of NaOH contains 1.6g of Oxygen
4.8g of SO2contains 2.4g of Oxygen
4.0 g of CO2contains 2.9g of Oxygen
2.8 g of CaO contains 0.8g of Oxygen
Question 9. Maximum number of electrons is present in
  1.    2.24 lit. of SO2 at S.T.P
  2.    0.2 moles of NH3
  3.    1.5 gm moles of oxygen
  4.    2 mole atoms of of sulphur
 Discuss Question
Answer: Option D. -> 2 mole atoms of of sulphur
:
D
Two moles of sulphur atoms have the maximum number of electrons.
No.ofelectrons=No.ofmoles×Avogadrono×No.ofelectronsinamolecule=2×N×16=32N
Question 10. No. of moles of water in 488.6 g of BaCl2.2H2O are (molecular weight of BaCl2.2H2O = 244.33)
  1.    2 moles
  2.    4 moles
  3.    3 moles
  4.    5 moles
 Discuss Question
Answer: Option B. -> 4 moles
:
B
No. of moles = WMW
1 mole contains 2 moles of water
2 moles contains 4 moles of water

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