Mole Concept(12th Grade > Chemistry ) Questions and Answers
Explanation:-
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
Explanation:-
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
Explanation:-
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
Explanation:-
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)
Explanation:-
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
Explanation:-
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