MUNISH KAKAR's INSTITUTE OF CHEMISTRY
MISCELLANEOUS MOLE CALCULATIONS & EMPIRICAL FORMULA WS#4
Q1.
From 510 mg of ammonia ( NH 3 ) sample, 6.02  10 21 molecules of ammonia are removed.
Calculate ;
Q2.
(a) Number of mole of ammonia removed
0.01 mole
(b) Number of mole of ammonia left
0.02 mole
Calculate the number of atoms of oxygen present in 88 g of CO 2. What would be the
weight of carbon monoxide having the same number of oxygen atoms?
(i) 0 atom = 4 × 6.02 × 1023
Q3.
Oxygen and nitrogen are present in a mixture in the ratio 1 : 4 by mass. Calculate the
ratio of number of molecules of N2 and O2
Q4.
32 : 7
How many years it would take to spend Avogadro number of rupees at the rate of one
million rupees per second.
Q5.
1.91 × 1010 years
Calculate the number of gram atoms and gram molecules in 2.54 mg of iodine (I 2).
(atomic mass I = 127 )
Q6.
(ii) 112 g
No. of atom of I = 2 × 10-5, No. of gram molecule = 1 × 10-5
Following is a crude but effective method for determination of magnitude of Avogadro
number by using stearic acid ( C18H36O2 ). When stearic acid ( molar mass = 284 g mol -1)
is added to water, it’s molecules collect at the surface and form a mono layer only. The
cross – sectional area of each stearic acid molecule has been measured to be 0.21 nm2 . In
one experiment, it is found that 7  10
5
g of stearic acid is needed to form monolayer
over water in a dish of diameter 20 cm. Based on these measurement, calculate the value
of avogardo's number.
Q7.
6.04 × 1023 (approx)
What is Empirical formula and Molecular formula ? How will you mathematically relate
Empirical formula and Molecular formula ?
Q8.
A compound of analysis was found to have the composition : sodium =14.31%,
sulphur=9.97%, oxygen=69.50%, hydrogen=6.22%.
Calculate the molecular formula of
the compound assuming that whole of hydrogen in the compound is present as water of
crystallization. Molecular mass of the compound is 322.
Q9.
Na2SO4.10H2O
5.0 g of a certain element X forms 10.0 g of its oxide having the formula X 4O6. The atomic
mass of X is
(a) 12.0 amu
(b) 24.0 amu
(c) 30.0 amu
(d) 32.0 amu
Q10. 0.5400 g of a metal X yields 1.020 g of its oxide X2O3. The number of moles of X is :
(a) 0.01
(b) 0.02
(c) 0.04
(d) 0.05
Q11. The phosphate of certain metal M is M3(PO4)2. Then, the correct formula of metal sulphate
would be:
(a) M2(SO4)3
(b) MSO4
(c) M3(SO4)2
(d) M2SO4
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Q12. Two elements X (atomic weight = 75) and Y (atomic weight = 16) combine to give a
compound having 75.8% of x. The formula of the compound is
(a) X3Y2
(b) X3Y
(c) XY2
(d) X2Y3
Q13. A compound made of two elements A and B is found to contain 25% A (atomic mass 12.5 )
and 75% B (atomic mass 37.5). The simplest formula of the compound is
(a) AB
(b) A2B2
(c) AB3
(d) A3B
Q14. A given sample of pure compound contains 9.81 gm of Zn, 1.8  1023 toms of chromium and
0.60 mole of oxygen atoms. What is the simplest formula ?
(a) ZnCr2O7
(b) ZnCr2O4
(c) ZnCrO4
(d) ZnCrO6
Q15. In a compound, C , H and N atoms are present in 9.33 : 1 : 1.55 by weight. Molecular
weight of the compound is 107. Molecular formula of compound is
(a) C7H9N
(b) C4H11N
(c) C3H7N
(d) C2H7N
Q16. A compound (60 g) on analysis gave C = 24 g, H = 4g, O= 32 g. Its empirical formula is:
(a) C2H2O2
(b) C2H2O2
(c) CH2O2
(d) CH2O
Q17. In a hydrocarbon, the mass ratio of hydrogen to carbon is 1 : 3, the empirical formula of
the hydrocarbon is
(a) CH
(b) CH2
(c) CH4
(d) CH3
Q18. The empirical formula of a compound is CH2O and its molecular weight is 120. The
molecular formula of the compound is
(a) C4H4O2
(b) C3H6O3
(c) C4H8O4
(d) CH2O
Q19. The hydrated salt Na2CO3.x H2O undergoes 63% loss in mass on heating and becomes
anhydrous. The value of x is ;
(a) 3
(b) 5
(c) 7
(d) 10
Q20. The simplest formula of a compound containing 50% of element X (atomic mass = 10
amu) and 50% of element Y (atomic mass = 20 amu) is ;
(a) XY
(b) X2Y
(c) XY3
(d) X2Y3
Q21. A mixture of O2 and gas “Y”(mol. wt. 80) in the mole ratio a : b has a mean molecular
weight 40. What would be mean molecular weight, if the gases are mixed in the ratio b : a
under identical conditions? (assuming gases are non-reacting) :
(a) 40
(b)48
(c)62
(d) 72
Q22. The number of moles corresponding to 8g Ca , 6.4 g of SO 2 and 6.72 L of CO2 are ;
a) 0.2 , 0.1 , 0.3
b) 0.1 , 0.2 , 0.3
Q23. In the following equation, x ( mole ) is ;
c) 5 , 10 , 3.33
d) none of these
x = 9.4g of phenol (C6H5OH) + 6.02  1023 molecules of phenol  0.2 mole of phenol x is ;
(a) 0.9 mol
(b) 9.2 g
(c) 0.1 mol
(d) 6.02  1023 molecules
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Q24. A sample of ammonium phosphate, (NH4)3PO4, contains 6 moles of hydrogen atoms. The
number of moles of oxygen atoms in the sample is:
(a) 1
(b) 2
(c) 4
(d)6
Q25. 1 mole CH4 contains:
(a) 6.02  1023 molecules of CH4
(b) 4  6.02 1023 atoms of H
(c) 6.02  1024 electrons in CH4
(d) All of these
Q26. Which of the following has the least mass ?
(a) 2g-atom of nitrogen ( A =14 )
(b) 3  1023 atoms of C
(c) 1 mole of S ( A = 32 )
(d) 7.0 g of Ag ( A =108 )
Q27. If an electron has mass 9.1  1031 Kg , the number of moles corresponding to 1 Kg electron
would be ;
a)
1
9.1  1031
b)
6.023  1023
1
c)
31
31
9.1  10
9.1  10  6.023  1023
d) 9.1  1031  6.023  1023
Q28. 11 grams of a gas occupy 5.6 litre of volume at S.T.P. The gas is:
(a) CO2
(b) N2O
(c) CO
(d) both a and b
Q29. The number of oxygen atoms in 4.4 g of CO2 is ;
(a) 1.2  1023
(b) 6  1022
(c) 6  1023
(d) 12  1023
Q30. The volume occupied by 4.4 g of CO2 at S.T.P. is:
(a) 22.4 L
(b) 2.24 L
(c) 0.224 L
(d) 0.1 L
Q31. The number of moles of oxygen in 1L of air containing 21% oxygen by volume, in standard
conditions is:
(a) 0.186 mol
(b) 0.21 mol
(c) 2.10 mol
(d) 0.0093 mol
Q32. The number of electrons in 22.4 L of hydrogen gas measured under STP condition are :
(a) 6.023  1023
(b) 12.046  1023
(c) 3.0115  1023
(d) zero
Q33. 2 g of oxygen contains number of atoms equal to that is:
(a) 0.5 g of hydrogen
(b) 4 g of sulphur
(c) 7 g of nitrogen
(d) 2.3 g of sodium
Q34. The largest number of molecules is in:
(a) 34 g of water
(b) 28 g of CO2
(c) 46 g of CH3OH
(d) 4 g of H2
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Q35. The number of water molecules in 1 litre of water is:( where NA is avogadro number )
(b) 10  1000
(a) 18
(c) NA
(d) 55.55 NA
Q36. The mass of 112 cm3 of CH4 gas at S.T.P. is:
(a) 0.16 g
(b) 0.8 g
(c) 0.08 g
(d) 1.6 g
Q37. If NA is Avogadro’s number, then number of valence electrons in 4.2 g of nitride ions (N 3-):
(a) 2.4 NA
(b) 4.2 NA
(c) 1.6 NA
(d) 3.2 NA
Q38. If 1021 molecules are removed from 200 mg of CO2, then the number of moles of CO2 left
are:
(a) 2.68  10-3
(b) 28.8  10-3
(c) 0.288  10-2
(d) 1.68  10-2
Q39. Rearrange the following (I to IV) in the order of increasing masses
I.
1 molecule of oxygen
II. 1 atom of nitrogen
1  10-10 g-molecular weight of oxygen
III.
(a) II < I < III < IV
(b) IV < III < II < I
IV. 1  10-10 g-atomic weight of copper
(c) II < III < I < IV
(d) none of these
Q40. Match the Column-I with Column-II:
Column-I
Column-II
1
1.6 g of methane
A
6.023  1023 molecules
2
3.2g of dioxygen
B
6.023  1023 protons.
3
22.4 L NH3
C
6.023  1023 electrons
4
1 gram molecule
D
6.02  10 22 molecules
1 = B, C, D
2=D
3=A
4=A
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