12th Grade > Chemistry
MOLE CONCEPT MCQs
Total Questions : 30
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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
:
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
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
:
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
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=0.01×1500×10−3(1mL=10−3L)
⇒ Molarity = 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=0.01×1500×10−3(1mL=10−3L)
⇒ Molarity = 0.02 M
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)
:
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)
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?
170.43×10−11g
:
B
No. of molecules required = 3×1010×102=3×1012
6×1023 molecules weight 342 g
3×1012 molecules weight?
170.43×10−11g
Answer: Option B. -> 200 ml
:
B
NaH2PO4 basicity = 2
No of moles ofNaH2PO4 =12120
ENaOH=ENaH2PO4
V×1M =12120×1000×2
V =200ml
:
B
NaH2PO4 basicity = 2
No of moles ofNaH2PO4 =12120
ENaOH=ENaH2PO4
V×1M =12120×1000×2
V =200ml
Answer: Option B. -> 1.26×1023 carbon atoms
:
B
∴1moleofC6H12O6has=6NAatomsofC∴0.35moleofC6H12O6has=6×0.35NAatomsofC=2.1NAatoms
=2.1×6.022×1023=1.26×1024 carbon atoms
:
B
∴1moleofC6H12O6has=6NAatomsofC∴0.35moleofC6H12O6has=6×0.35NAatomsofC=2.1NAatoms
=2.1×6.022×1023=1.26×1024 carbon atoms
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
:
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
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
:
D
Two moles of sulphur atoms have the maximum number of electrons.
No.ofelectrons=No.ofmoles×Avogadrono×No.ofelectronsinamolecule=2×N×16=32N
Answer: Option B. -> 4 moles
:
B
No. of moles = WMW
1 mole contains 2 moles of water
2 moles contains 4 moles of water
:
B
No. of moles = WMW
1 mole contains 2 moles of water
2 moles contains 4 moles of water