Mole Concept(11th And 12th > Chemistry ) Questions and Answers
Explanation:-
Answer: Option B. -> 4.48:
B
The question gives weight of reactants and asks the volume of product. So, it is a weight-volume problem. Start by balancing the reaction
CaCO3 → CaO + CO2(g) ↑
∴ 1 mole of CaCO3 gives 1 mole of CO2
From Avagadro's Law, 1 mole of CO2 at NTP occupies 22.4L
∴ 1 mole of CaCO3 = 22.4 L of CO2
100g = 22.4L of CO2
∴ 20g = 20 × 22.4100 = 4.48 L of CO2
Explanation:-
Answer: Option B. -> 12.011 u:
B
Average atomic mass is the weighted average of atomic mass of all the isotopes.
Contribution by each isotope = (Relative abundance) (Atomic mass)
Contribution by
126C = (0.98892)(12)/100 = 11.86704
136C = (0.01108)(13.00335)/100 = 0.14407712
146C = (2 × 10−10)(14.003170/100 = 28.00634 × 10−12
∴ Average atomic mass = sum of all contribution = 12.011 u
Explanation:-
Answer: Option A. -> 1±0.1%;10±0.01%;100 ±0.001%:
A
(a) 1±0.1% (b) 10±0.01% (c) 100±0.001%
The uncertainty in measurement is expressed in terms of percentage by putting ± sign before it, e.g., 150±1%, et. Smaller the quantity to be measured, greater is the percentage uncertainty, and the instrument should be more precise for the measurement of smaller quantities.
1mg=11000g11000×100=110=0.11±0.1%
Explanation:-
Answer: Option B. -> 11.39:
B
The question talks about reaction between Fe and CuSO4. Observe that weights of the reactants are given and the weight of a product is asked. Thus, it is a weight-weight stoichiometry problem.
In all such stoichiometry problems, start by balancing the reaction
Fe + CuSO4 → FeSO4 + Cu
From the equation, 1 mole of Fe = 1 mole of Cu
∴ 55.85g ≈ 63.6g of Cu
∴ 10g of Fe will displace 11.39g of Cu if available.
The solution has 70g of CuSO4
1 mole CuSO4 has 63.6g of Cu
i.e., [63.6 + 32 + 16(4)]g of CuSO4 = 63.6g of Cu
159.6g of CuSO4= 63.6g of Cu
70g of CuSO4= 27.76g of Cu
∴ 10 g = 10 × 63.655.85 = 11.39 g of Cu
∴ In the CuSO4 solution, 27.76g of Cu is present of which 11.39g of Cu will be precipitated out by the addition of 10g of Fe.
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. -> 50 ml:
B
2NaOH +H2SO4⟶ Na2SO4 + 2H2O
0.0025 moles ofH2SO4 will neutralise 0.005 moles NaOH.
So for 0.01 V moles H2SO4 there will be 0.01v-0.0025 moles H2SO4 left unused in the total volume(v+50ml) of solution .
The concentration here will be
0.01−0.0025v+50×1000 = 0.05M
100V−2.5V+50 =0.05
100V−2.5V+50= 2.5+2.5
99.95V= 5
V= 0.05002L
V= 50.02ML
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 =Given weightMolecular weight(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