Atoms, Molecules and Chemical Arithmetic NIOS Terminal Questions With Answer
National Institute of Open Schooling (NIOS)
Sr. Secondary, Module-1
Lession-1: Atoms, Molecules and Chemical Arithmatics Terminal Questions With Answer
Terminal QUESTIONS with Answer
1. How many atoms are present in a piece of iron that has a mass of 65.0 gm ?
(atomic mass Fe = 55.9 amu)
Answwer:1 mole of Fe contains 6.022 x 1023 atoms
i.e. 55.9 gm of Fe contains 6.022 × 1023 atoms
∴ 65 gm of Fe contains = 6.022 × 1023 × 65 / 55.9 = 7.0023 × 1023
So, 65 gm of Fe contains 7.0023 × 1023 atoms of Fe.
2. A piece of phosphorus has a mass of 99.2 gm. How many moles of phophorus, P are present in it? (atomic mass P = 31.0 amu)
Answer:Molecular wt. of P4 = 4 × 31.0= 124 gm.
Now, 124 gm of P contains I mole
∴ 99.2 gm of P4 contains = 1 × 99.2/124 = 0.80
∴ 99.2 gm of P contains 0.80 moles of phosphorus.
3. Mass of 8.46 × 1024 atoms of fluorine is 266.95 g. Calculate the atomic mass of flourine.
Answer:Atomic mass = Mass of 6.022 × 1023 atoms
Now, 8.46 x 1024 atoms has mass 266.95 g
∴ 6.022 × 1023 atoms has mass
6.022 × 1023 × 266.95 / 8.46 × 1024 = 19
So the atomic mass of flourine is 19 amu.
4. A sample of magnesium consists of 1.92 x 1022 Mg atoms. What is the mass of sample in grams? (atomic mass = 24.3 amu)
Answer:Mass of 6.022 × 1023 atoms is 24.3 gm.
Mass of 1.92 × 1022 atoms = 24.3 × 1.92 × 1022 / 6.022 × 1023 = 0.77 gm
So the mass of sample is 0.77 gm.
5. Calculate the molar mass in g mol-1 for each of the following:
(i) Sodium hydroxide, NaOH
(ii) Copper sulphate, CuSO4.5H2O
(iii) Sodium Carbonate, Na2CO3.10H2O
Answer:(i) Molar mass of NaOH in g mol-1 is 23 + 16 + 1 = 40 amu
(ii) Molar mass of CuSO4.5H2O = 65.5 + 32 + (4 × 16) + 5 (18) = 251.5 amu
(iii) Molar mass of Na2CO3.10H2O = 2 × 23 + 12 + (3 × 16) + (10 × 18) = 286 amu
6. For 150 gram sample of phosphorus trichloride PCl3, calculate each of the following:
(i) Mass of one PCI3 molecule.
(ii) The no. of moles of PCI3 and Cl in sample.
(iii) The no. of grams of Cl atoms in the sample.
(iv) The no. of molecules of PCI3 in the sample.
Answer:(i) Molecular mass of PC13 = 137.5
i.e. Mass of 6.022 x 1023 molecules is 137.5 gm
∴ Mass of 1 molecule = 137.5 / 6.022 × 1023 = 22.83 x 10-23 gm = 22.83 x 10-26 kg
Mass of 1 molecule of PCl3 is 22.83 × 10-26 kg
(ii) Number of moles of PCI3 = 150 / 137.5 = 1.09
No. of moles of Cl in sample = 3 × 1.09 = 3.27
(iii) In PC13 the ratio of mass of CI: PCI3 = (3 × 35.5) : 137.5
= 106.5 : 137.5
= weight of Cl : 150
∴ Weight of Cl = 150 × 106.5 / 137.5 = 116.18 gm
So, the weight of Cl in 150gm of PCI3 = 116.18 gm
(iv) Number of molecules of PCI3 in sample = 1.09 × 6.022 x 1023
= 6.56 x 1023 molecules of PCI3
7. Find out the mass of C-12 that would contain 1 × 1019 atoms.
Answer:Mass of 6.022 × 1023 atoms = 12gm
So the mass of 1 x 1019 atoms = 12 × 1019 / 6.022 × 1023 = 1.99 x 10-4
Therefore, the mass of sample is 1.99 x 10-4 gm.
8. How many atoms are present in 100 g sample of C-12 atom?
Answer:12gm sample contains 6.022 × 1023 atom
Therefor, 100 g sample will contain 100 × 6.022 × 1023 / 12 atoms
= 8.33 × 6.022 × 1023 = 50.16 x 1023 atoms
Therefore, 100 gm sample contains 50.16 × 1023 atoms.
9. How many moles of CaCO3 would weigh 5 gm?
Answer:Molecular weight of CaCO3 = 40 + 12 + 48 = 100
i.e. 100 gm of CaCO3 contains 1 mole
therefore, 5 gm of CaCO3 contain 0.05 mole.
10. If you require 1 × 1023 molecules of nitrogen for the reaction N2 + 3H2 → 2NH3.
(i) What is the mass (in gm) of N2 required?
(ii) How many moles of NH3 would be formed in the above reaction from 1.0 × 1023 molecules of N2?
(iii) What volume would NH3 gas formed in (ii), occupy at STP?
Answer:(i) 6.022 × 1023 molecules weight 28 gm
Therefore, 1023 molecules weight = 28 / 6.022 = 4.65 gm
So the weight of 1023 molecules is 4.65 gm
(ii) Moles of NH3 formed = 2 x moles of nitrogen = (2 x 4.65) /28 = 0.33
So the number of moles NH3 formed is 0.33.
(iii) 1 mole of substance has volume 22.7 litre at STP.
Therefore, 0.33 mole will occupy = 22.7 × 0.33 = 7.49 liter
Therefor, NH3 so formed will occupy 7.49 litre at STP.
11. Write empirical formulae of following compounds: CO, Na2S, C4H10, H2O2, KCI
Answer:| Compound | Emperical Formula |
|---|---|
| CO | CO |
| Na2SO3 | Na2SO3 |
| C4H10 | C2H5 |
| H2O2 | HO |
| KCI | KCI |
12. The empirical formula of glucose is CH2O which has a formula mass of 30 amu. If the molecular mass of glucose is 180 amu. Determine the molecular formula of glucose.
Answer:We know that the relation between molecular formula and empirical formula is given by-
molecular formula = n × empirical formula
∴ 180 = n × 30
or, n = 180 / 30 = 60
molecular formula = n × empirical formula
or, molecular formula = 6 × CH2O = C6H12O6
So the molecular formula of glucose is C6H12O6
13. What is the ratio of masses of oxygen that are combined with 1.0 gm of nitrogen in the compound NO and N2O3?
Answer:For NO
Mass of oxygen per mole of NO = 16 gm
Mass of nitrogen per mole of NO = 14 gm
So, Mass of oxygen in NO per 1.0 gm of nitrogen = 16/14
For N2O3
Mass of oxygen per mole of N2O3 = 16 × 3 = 48 gm
Mass of nitrogen per mole of N2O3 = 14 × 2 = 28 gm
So, Mass of oxygen in N2O3 per 1.0 gm of nitrogen = 48/28 = 2/14
Thus, Required ratio = (16/14) / (24/14) = 16 / 24
14. A compound containing sulphur and oxygen, on analysis reveals that it contains 50.1% sulphur and 49.9% oxygen by mass. What is the simplest formula of the compound?
Answer:Let, the formula required be SxOy
So the molar Mass of SxOy = 32.06x + 16y
Mass of Sulphur per mol of SxOy = x × 32.06
So, % of Sulphur per mole of SxOy = [(x × 32.06) / (32.06x + 16y)] × 100
or, 50.1 = [(x × 32.06) / (32.06x + 16y)] × 100
or, 50.1 × (32.06x + 16y) = 3206x
or, 606.21x + 801.60y = 3206x
or, 801.60y = 1599.79y
or, x/y = 801.60 / 1599.79
x:y = 1:2
Thus, simplest required formula or empirical formula = SO2
15. Hydrocarbons are organic compounds composed of hydrogen and carbon. A 0.1647 g sample of pure hydrocarbon on burning in a combustion tube produced 0.5694 g of CO2 and 0.0845 g of H2O. Determine the percentage of these elements in hydrocarbons?
Answer:Sol. Reaction will be
Hydrocarbon + Oxygen → CO2 + H2O
0.1647 g 0.5694g 0.0845 g
Let empirical formula of hydrocarbon be CxHy
Balanced Reaction is-
Cx Hy + x O2 + y/4 O2 → xCO2 + (y/2) H2O
Molecular Mass of CO2 = 12 + (16 × 2) = 44 gm
and Molecular Mass of H2O = (1 × 2)+16 = 18 gm
Mole of CO2 in 0.5694 gm = 0.5694/44 = 0.0129
Moles of H2O in 0.0845 gm = 0.0845/18 = 0.00469
Therefore, x : (y/2) = 0.01:0.08 = 1:8
or, x:y = 1:4
Thus, emperical formula of hydrocarbon = CH8
Now, % of carbon in CH8 = [12/(12+8)] × 100 = 60%
and % of hydrogen in CH8 = [1 × 8/(12+8)] × 100 = 40%
