Lakshya Education MCQs

Question: IsotopeRelative Abudance(%)Atomic Mass(amu)12C98.8921213C1.10813.0033514C2×101014.00317

From the data given in the table, calculate the average atomic mass of C
Options:
A.13 u
B.12.011 u
C.12.001 u
D.12.0011 u
Answer: Option B
: 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 × 1010)(14.003170/100 = 28.00634× 1012 Average atomic mass = sum of all contribution = 12.011 u

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Question 1. 10g of Fe powder was mixed with a CuSO4 solution which contained 70g of CuSO4. What is the weight of the copper obtained?

[Fe = 55.85g        Cu = 63.6g]
  1.    10.39
  2.    11.39
  3.    12.39
  4.    13.39
Answer: Option B
: 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.
Question 2. Avogadro’s number represents the number of atoms in
  1.    12g of C -12
  2.    32g of sulphur
  3.    32g of oxygen
  4.    12.7g of iodine
Answer: Option A
: A

It is the total number of particles (atoms, molecules) present in C12 isotope
Question 3. What volume of 0.10 MH2SO4  must be added to 50 ml of a 0.10 M NaOH solution to make a solution in which the molarity of theH2SO4 is 0.050 M?
  1.    400 ml
  2.    50 ml
  3.    100 ml
  4.    150 ml
Answer: Option B
: B

2NaOH +H2SO4Na2SO4 + 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 0.010.0025v+50×1000 = 0.05M 100V2.5V+50=0.05 100V2.5V+50= 2.5+2.5 99.95V= 5 V= 0.05002L V= 50.02ML
Question 4. How many years it would take to spend Avogadro's number of rupees at the rate of 1 million rupees in one second?
  1.    19.098×1019 years
  2.    19.098 years
  3.    19.098×109 years
  4.    None of these
Answer: Option C
: C

106 rupees are spent in 1sec. 6.022×1023 rupees are spent in
=(1×6.022×1023)(106×60×60×24×365)years
=19.098×109year

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