Subjective Assignment Moleconcept- 3 At. Mass K = 39; N = 14; O = 16; C = 12, Na = 23

1. Calculate mass of 0.1 mole of KNO3,1023 molecule of CH4 ,112 ml of H2 at STP.

Sol. Mass = 0.1 × 101 =10.1 gm KNO3

= 6.21010022.6

16 2323 =×

× gm CH4

= 71222400

2× = 0.01

2. Calculate volume at STP occupied by 14gm of NO2 gas. Sol. 46 gm = 22.4 lit

14 gm = 1446

4.22× = 6.81 lit

3. Calculate the mass of Na which contains same No. of atoms as are present in 15 gm of Ca.

Sol. Mole of Ca = 4015 ; same mole of Na =

4015

Mass of Na = 234015

× = 8.625 gm

4. What volume is occupied at NTP (a) 0 .2 mole of NH3 (b) 6.023 ×

1021 molecule of oxygen. Sol. (a) 0.2 × 22.4 = 4.48 lit NH3

(b) = 2123 10023.6

106.022lit 4.22

×××

= 0.224 lit.

5. 1 atom of an element X weights 6.644 × 10–23 gm . Calculate No. of gm atom in 80kg of it.

Sol. Mass of Avagadro No. = 6.644 × 10-23 × 6.0233 × 1023 = 40 gm 40 gm = 1 mole

80 × 103 gm = 40

1080 3× = 2000 mole

6. Calculate the no. of gold atoms in 300mg of a gold ring of 20 carot

gold.[At. Mass =197; pure gold = 24 carot.] Sol. Percentage purity = 100

2420

× = 83.333%

Mass of gold = 0.3 × 100

33.83 = 0.250 gm

No. of atoms = 2310022.6197250.0

××

= 7.64 × 1020

7. What mass in kg of K2O contains the same number of moles of K-atoms as are present in 10 kg of KCl.

Sol. Moles of KCl = 228.1345.74

000,10=

Moles of K2O = 2228.134

Mass of K2O = 942228.134

× = 6308.7 gm

= 6.307 Kg

8. Calculate amount of N in 192 gm sample of TNT. [M.Wt. = 227] & Molecular formula

Sol. 192

22742

×= = 35.5

CH3

NO2

NO2

O2N

9. Calculate total no. of e- and proton in 3.2 gm of CH4. Sol. Mole of CH4 =

322.3 = 0.1 mole

No. of Molecules = 0.1 × 6.022 × 1023

e– + p = 20 × 0.1 × 6 × 1023 = 12 × 1023

10. If the atomic mass of carbon is set at 150 u. Calculate value of

Avogadros Number. Sol. 12 u = 6.023 × 1023

150 u = 15012

10023.6 23×

× = 7.33 × 1024

11. How many moles of Ba, Cl & H2O are present in 122 gm BaCl2.2H2O [M.Wt. of BaCl2.2H2O = 244]

Sol. 21mole Ba, 1 mole Cl, 1 mole H2O

12. Calculate number of atoms in

(a) 2 gm He (b) 4 amu He (c) 20 mole CaCO3 Sol. (a) 3 × 1023 (b) 1 (c) 20 × 6.023 × 1023 × 5 = 6.023 × 1025

14. Assume the human body contain 80% water. Calculate the number of molecule of water that are in the body of a person who has a mass of 65 kg.

Sol. Wt. of water in human body = 6510080

× = 52 kg

18 gm H2O has = 6.022 × 1023 molecules

52 × 103 gm = 323

105218

10022.6××

× = 17.39 × 1026

15. If it requires 1 second to count one wheat grain. Calculate the time

(in years) required to count half mole of wheat grains.

Sol. 21mole of grain = 2310023.6

21

××

Time require = 3 × 1023 second

year365246060

103 23

××××

= = 9.54 × 1015 year

15. The atom of an element weighs. 6.645 × 10–23 gm. Name the element.

Sol. Mass of Avogadro Number = Atomic mass = 6.022 × 1023 × 6.645 × 10–23 gm = 40 gm so it is ‘Ca’ 16. 0.01 mole of a compound weighs one gram. Calculate the

molecular mass of the compound. Sol. 100 amu. 17. What is the number of molecule of CO2 that have 8 gm O2. Sol. moles in 8gm O2 =

41

328=

therefore No. of CO2 = 6.022 × 1023 × 41

=1.5 × 1023

18. How many mol are present in 21.44 gm Ti S1.85 [S = 32; Ti = 47.9]

Sol. M. Mass of TiS1.85 = 47.9 + 32 ×1.85 = 107.10 107.10 gm have = 1 mole

21.44 gm = 44.2110.107

1× = 0.2002 mol.

19. The ratio of mass of an atom of oxygen to an atom of C is 3:4 if

the m.wt. of carbon is 12gm/mole. What is the mass of 1 molecule of CO2.

Sol. O : C M.Wt of CO2 = 12 + 18 3 : 4 = 30

9gm 12gm Mass of 1 molecule = 2310022.630×

gm

= 4.98 × 10–23gm

20. Sample of CaCl2 contain 6.022 × 1020 Ca+2. Calculate (i) No. of formula units (ii) Mass of sample (iii) No. of Cl– ion

Sol. No. of formula unit = 6 × 1020

Mass of sample = 111106106

23

20×

×

× = 0.111kg

No. of Cl– ion = 2 × 6 × 1020 = 12 × 1020

21. An element X has the following isotopic composition 200X = 90%;

199X = 8% ; 202X = 2% What will be average atomic mass of naturally occurring element X.

Sol. Av. Mass = 100

2022819920090 ×+×+× = 199 amu

22. 10 lit. O2 gas is reacted with 30 lit. of CO at STP. What will be volume of each gas present at the end of reaction.

Sol. Reaction

lit. 20

2

lit. 20lit. 2

lit. 10lit. 1

2 CO CO O21

→+

CO = 10 lit; CO2 = 20 lit ; O2 = 0 lit. 23. Suppose that Si – 28 (( )Si28

14 is taken as the standard for expressing atomic masses & assigned an atomic mass of 10.00 amu. Estimate the molar mass of lithium nitride (Li3N) (if atomic mass of Li is 7 amu & N = 14 amu at C – 12 taken as standard)

Sol. 12.5

24. Consider an isotope yttrium y9039 . This isotope is incorporated into cancer-

seeking antibodies so that the cancer can be irradiated by the yttrium & destroyed. How many neutrons are in (a) 25 atoms of yttrium (b) 0.25 mol of yttrium (c) one nanogram (10-9 gm) of yttrium?

Sol. (a) 1275 neutrons (b) 7.68 × 1024 (c) 3.4 × 1014

25. Simplify for x mass of 2

N0 atoms of 12C +1.2 g of 12C – 0.1 N0 atoms of 12C

= x mol of 12C. Sol. We are to derive the result in terms of mol.

Mass of 2

N0 atoms of 12C = g 6N12

2N

00 =× = 0.5 mol

1.2 g of 12C = mol 1.012

2.1=

N0 atom of 12C = 1 mol

∴ 0.1 N0 atoms12C = 0.1 mol

Hence x = 0.5 + 0.1 – 0.1 = 0.5 mol

26. A sample of MgCO3 is 50% pure. If on heating 4.0 g of MgO and 4.4 g of CO2 are formed, calculate amount of MgCO3 taken (Mg = 24, C = 12, O = 16)

Sol.

g 442

g 40g 843 )g(CO )s(MgO)S(MgCO +⎯→⎯Δ

Amount of products = 4.0 + 4.4 = 8.4 g

∴ Amount of reactant (MgCO3) = 8.4 g

But MgCO3 is 50% pure, hence amount taken = 8.4 × 50

100

= 16.8 g

27. In the decomposition of impure KClO3, 4.9 g of it gave 1.49 g of KCl and 0.03 mol of O2 gas. What is per cent purity of KClO3?

Sol. (g)3O )s(KCl2)s(KClO2g 0.96 mol 0.03

2g 49.1(impure) g 9.4

3=

Δ +⎯→⎯

By law of conservation of mass, amount of reactants and products are equal = 1.49 + 0.96 = 2.45 g

• But reactant (KClO3) taken = 4.9 g

• Thus, percent purity = 1009.4

45.2× = 50%

28. Methane and ethane are both constituents of natural gas. A sample of methane contains 11.40 g of carbon and 3.80 g of hydrogen, whereas a sample of ethane contains 4.47 g of carbon and 1.118 g of hydrogen. Show that the two substances obey the law of multiple proportions.

Sol. First, find the C : H mass ratio in each compound.

Methane : C : H mass ratio = 00.3gH 80.3

C g40.11=

Ethane : C : H mass ratio = 00.4gH 118.1C g47.4=

Thus, two C : H ratios are clearly small, whole-number multiples of each other.

43

ethane in ratio mass H:Cmethane in ratio mass H:C

=

29. A divalent cation is isoelectronic of CO2 and has (Z + 2) neutrons. What is ionic mass of divalent cation?

Sol. A divalent cation (M2+) has as many as electrons as CO2 = 6 + 16 = 22 Hence, (Z) atomic number of M2+ = 22 + 2 = 24 or number of protons (P) in M2+ = 24 number of neutrons (N) in M2+ = 24

∴ ionic mass of M2+ = P + N = 24 + 26 = 50

30. 105 B and B11

5 are two isotopes of boron. If average mass number is 10.2, what is the percentage of each?

Sol. Let percentage of 105 B = x

then percentage of B115 = (100 – x)

Average mass number A = 21

2211XX

XAXA++

∴ x100x

)x100(11x102.10−+−+

=

100

x111100x102.10 −+=

x = 80

Thus, percentage of B105 = 80%

and percentage of B115 = 20%