Answers Assignment No 2 - Some basic Concepts of Chemistry |Class 11th

Answers

Assignment No – 2

Some Basic Concepts of Chemistry

---

Ans 1. Method to convert units from one system to other is called unit factor method.

93 million miles = 93 x 10⁶ miles

1 mile = 1.60934 km = 1.60934 x 10³ m

Ans 2. Atomic mass of an element expressed in grams is the gram atomic mass

Number of silver atoms = 1 million = 1 x 106

Mass of one million Ag atoms = 1.79 x 10¹⁶g

Mass of 6.023 x 10²³ atoms of Silver = 1.79 x 10¹⁶g x 6.023 x 10²³

                                                             1 x 10⁶

 = 107.8 g

Ans 3. Law of Definite Proportions: According to this law a chemical compound

always consists of the same elements combined together in the same ratio,

irrespective of the method of preparation or the source from where it is taken.

Ans 4. (a)NO₂ (b) CH₂O (c) H₂O (d) HO

Ans 5. Average atomic mass = 34.97 x 0.755 +36.97 x 0.245 = 35.46 u

Ans 6. Molar mass of Na₂CO₃ = 2 x 23 +12 + 3 x 16 = 106 g / mol

0.50 mol Na₂CO₃ means 0.50 x 106 = 53 g

0.50 M Na₂CO₃ means 0.50 mol i.e. 53 g of Na₂CO₃ are present in I L of the

solution.

Ans 7. Molecular formula of ethanol is: C₂H₅OH

Molar mass of ethanol is: (212.01 + 61.008 + 16.00) g = 46.068 g

Mass per cent of carbon = (24.02g / 46.068g) ×100 = 52.14%

Mass per cent of hydrogen = (6.048g / 46.068g) × 100 = 13.13%

Mass per cent of oxygen = (16.00 g / 46.068g) ×100 = 34.73%

Ans 8. The number of moles of solute dissolved per litre (dm³) of the solution is called molarity.

Since molarity (M) = No. of moles of solute /Volume of solution in litres

= (Mass of NaOH / Molar Mass of NaOH) /0.250 L

= (4 g / 40 g 0.1 mol) / 0.250L = 0.1 mol / 0.250 L

= 0.4 mol L^-1  = 0.4 M

Ans 9. (a) 4.8 x 10^-3                              (b) 2.3 x 10⁵                      (c) 2.0 x 10²

Ans 10. (i) 1 mole of Ar = 6.022 × 1023 atoms of Ar

52 mol of Ar = 52 × 6.022 × 10²³ atoms of Ar

= 3.131 × 10²⁵ atoms of Ar

(ii) 1 atom of He = 4 u of He        

Or

4 u of He = 1 atom of He

1 u of He = 1/4 atom of He

52 u of He = 52/4 atom of He

= 13 atoms of He

(iii) Molar mass of He = 4 g/mol

4 g of He contains = 6.022 × 10²³ atoms of He

52 g of He contains = 6.022 × 10²³ x 52 = 78.286 x 10²³ atoms of He

    4

Ans 11. An empirical formula represents the simplest whole number ration of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

The empirical formula of the above compound is CH₂Cl.

empirical formula mass is 12 + (1 x 2) + 35.5 = 49.5

n= Molecular Mass / Empirical Formula Mass = 98.96 / 49.5 = 2

Hence molecular formula is C₂H₄Cl₂