# 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 10

^{23}atoms

i.e. 55.9 gm of Fe contains 6.022 × 10

^{23}atoms

∴ 65 gm of Fe contains = 6.022 × 10

^{23}× 65 / 55.9 = 7.0023 × 10

^{23}

So, 65 gm of Fe contains 7.0023 × 10

^{23}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 P

_{4}= 4 × 31.0= 124 gm.

Now, 124 gm of P contains I mole

∴ 99.2 gm of P

_{4}contains = 1 × 99.2/124 = 0.80

∴ 99.2 gm of P contains 0.80 moles of phosphorus.

### 3. Mass of 8.46 × 10^{24} atoms of fluorine is 266.95 g. Calculate the atomic mass of flourine.

Answer:Atomic mass = Mass of 6.022 × 10

^{23}atoms

Now, 8.46 x 10

^{24}atoms has mass 266.95 g

∴ 6.022 × 10

^{23}atoms has mass

6.022 × 10

^{23}× 266.95 / 8.46 × 10

^{24}= 19

So the atomic mass of flourine is 19 amu.

### 4. A sample of magnesium consists of 1.92 x 10^{22} Mg atoms. What is the mass of sample in grams? (atomic mass = 24.3 amu)

Answer:Mass of 6.022 × 10

^{23}atoms is 24.3 gm.

Mass of 1.92 × 10

^{22}atoms = 24.3 × 1.92 × 10

^{22}/ 6.022 × 10

^{23}= 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, CuSO_{4}.5H_{2}O

(iii) Sodium Carbonate, Na_{2}CO_{3}.10H_{2}O

Answer:(i) Molar mass of NaOH in g mol

^{-1}is 23 + 16 + 1 = 40 amu

(ii) Molar mass of CuSO

_{4}.5H

_{2}O = 65.5 + 32 + (4 × 16) + 5 (18) = 251.5 amu

(iii) Molar mass of Na

_{2}CO

_{3}.10H

_{2}O = 2 × 23 + 12 + (3 × 16) + (10 × 18) = 286 amu

### 6. For 150 gram sample of phosphorus trichloride PCl_{3}, calculate each of the following:

(i) Mass of one PCI_{3} molecule.

(ii) The no. of moles of PCI_{3} and Cl in sample.

(iii) The no. of grams of Cl atoms in the sample.

(iv) The no. of molecules of PCI_{3} in the sample.

Answer:(i) Molecular mass of PC1

_{3}= 137.5

i.e. Mass of 6.022 x 10

^{23}molecules is 137.5 gm

∴ Mass of 1 molecule = 137.5 / 6.022 × 10

^{23}= 22.83 x 10

^{-23}gm = 22.83 x 10

^{-26}kg

Mass of 1 molecule of PCl

_{3}is 22.83 × 10

^{-26}kg

(ii) Number of moles of PCI

_{3}= 150 / 137.5 = 1.09

No. of moles of Cl in sample = 3 × 1.09 = 3.27

(iii) In PC1

_{3}the ratio of mass of CI: PCI

_{3}= (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 PCI

_{3}= 116.18 gm

(iv) Number of molecules of PCI

_{3}in sample = 1.09 × 6.022 x 10

^{23}

= 6.56 x 10

^{23}molecules of PCI

_{3}

### 7. Find out the mass of C-12 that would contain 1 × 10^{19} atoms.

Answer:Mass of 6.022 × 10

^{23}atoms = 12gm

So the mass of 1 x 10

^{19}atoms = 12 × 10

^{19}/ 6.022 × 10

^{23}= 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 × 10

^{23}atom

Therefor, 100 g sample will contain 100 × 6.022 × 10

^{23}/ 12 atoms

= 8.33 × 6.022 × 10

^{23}= 50.16 x 10

^{23}atoms

Therefore, 100 gm sample contains 50.16 × 10

^{23}atoms.

### 9. How many moles of CaCO_{3} would weigh 5 gm?

Answer:Molecular weight of CaCO

_{3}= 40 + 12 + 48 = 100

i.e. 100 gm of CaCO

_{3}contains 1 mole

therefore, 5 gm of CaCO

_{3}contain 0.05 mole.

### 10. If you require 1 × 10^{23} molecules of nitrogen for the reaction N_{2} + 3H_{2} → 2NH_{3}.

(i) What is the mass (in gm) of N_{2} required?

(ii) How many moles of NH_{3} would be formed in the above reaction from 1.0 × 10^{23} molecules of N_{2}?

(iii) What volume would NH_{3} gas formed in (ii), occupy at STP?

Answer:(i) 6.022 × 10

^{23}molecules weight 28 gm

Therefore, 10

^{23}molecules weight = 28 / 6.022 = 4.65 gm

So the weight of 10

^{23}molecules is 4.65 gm

(ii) Moles of NH

_{3}formed = 2 x moles of nitrogen = (2 x 4.65) /28 = 0.33

So the number of moles NH

_{3}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, NH

_{3}so formed will occupy 7.49 litre at STP.

### 11. Write empirical formulae of following compounds:
CO, Na_{2}S_{}, C_{4}H_{10}, H_{2}O_{2}, KCI

Answer:Compound | Emperical Formula |
---|---|

CO | CO |

Na_{2}SO_{3} | Na_{2}SO_{3} |

C_{4}H_{10} | C_{2}H_{5} |

H_{2}O_{2} | HO |

KCI | KCI |

### 12. The empirical formula of glucose is CH_{2}O 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 × CH

_{2}O = C

_{6}H

_{12}O

_{6}

So the molecular formula of glucose is C

_{6}H

_{12}O

_{6}

### 13. What is the ratio of masses of oxygen that are combined with 1.0 gm of nitrogen in the compound NO and N_{2}O_{3}?

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 N

_{2}O

_{3}

Mass of oxygen per mole of N

_{2}O

_{3}= 16 × 3 = 48 gm

Mass of nitrogen per mole of N

_{2}O

_{3}= 14 × 2 = 28 gm

So, Mass of oxygen in N

_{2}O

_{3}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 S

_{x}O

_{y}

So the molar Mass of S

_{x}O

_{y}= 32.06x + 16y

Mass of Sulphur per mol of S

_{x}O

_{y}= x × 32.06

So, % of Sulphur per mole of S

_{x}O

_{y}= [(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 = SO

_{2}

### 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 CO_{2} and 0.0845 g of H_{2}O. Determine the percentage of these elements in hydrocarbons?

Answer:Sol. Reaction will be

Hydrocarbon + Oxygen → CO

_{2}+ H

_{2}O

0.1647 g 0.5694g 0.0845 g

Let empirical formula of hydrocarbon be C

_{x}H

_{y}

Balanced Reaction is-

C

_{x}H

_{y}+ x O

_{2}+ y/4 O

_{2}→ xCO

_{2}+ (y/2) H

_{2}O

Molecular Mass of CO

_{2}= 12 + (16 × 2) = 44 gm

and Molecular Mass of H

_{2}O = (1 × 2)+16 = 18 gm

Mole of CO

_{2}in 0.5694 gm = 0.5694/44 = 0.0129

Moles of H

_{2}O 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 = CH

_{8}

Now, % of carbon in CH

_{8}= [12/(12+8)] × 100 = 60%

and % of hydrogen in CH

_{8}= [1 × 8/(12+8)] × 100 = 40%