Some Basic Concepts of Chemistry

1.15. micro = 10^–6, deca = 10, mega = 10⁶, giga = 10⁹, femto = 10^–15

1.14. S.I. unit of mass is kilogram (kg).

It is defined as the mass of platinum-iridium block stored at international bureau of weights and measures in France.

1.13. Pressure= Force/Area

But weight = m x g, where m = mass (in kg) and g = 9.8 m/s²

Therefore, Pressure = 1034 g cm^-2 x 9.8 m/s²

= 1034 x 10^-3 kg x (100 m)² x 9.8 m/s²

= 101332 Pa

= 1.01332 x 10⁵ Pa

1.12. Density of methanol = 0.793 kg/L, molar mass of methanol (CH₃OH) = 32g/mol = 0.032 kg/mol

V₁ =? , V₂ = 2.5 L, M₂ = 0.25 M

We can apply the formula of

M₁V₁ = M₂V₂

Or V₁ = M₂V₂/M₁

Substituting M₁ = density / molar mass, we get

M₁ = 0.793/0.032 = 24.78

V₁ = 0.25 x 2.5 / 24.78 = 0.02522 L = 25.22 mL

1.11. Molar mass of sugar = (12 x 12) + (22 x 1) + (11 x 16) =342 g/mol

No. of moles in 352 g of sugar = 1 mol

No. of moles in 20 g = 20 x 1/352 = 0.0585 mol

Therefore, molar concentration = moles of solute / volume of solution in L = 0.0585 / 2 = 0.0293 mol/L

1.10. (i) 1 mole of C₂H₆ contains 2 moles of carbon atoms

     Therefore, no of moles of C atoms in 3 moles of C₂H₆ = 6 moles
(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms
      Therefore, no of moles of  H atoms in 3 moles of C₂H₆ = 18 moles

(iii) 1 mole of C₂H₆ contains 6.02 x 10²³ molecules

      Therefore, 3 moles of C₂H₆ will contain ethane molecules = 3 x 6.02 x 10²³ molecules= 18.06 x 10²³ molecules

1.9. Average atomic mass = (Fractional abundance of ³⁵Cl x molar mass of ³⁵Cl) + (Fractional abundance of ³⁷Cl x molar mass of ³⁷Cl)

= (75.77/100 x 34.9689) + (24.23/100 x 36.9659)

= 26.4959 + 8.9568

= 35.4527

1.8. Mass percentage of Fe = 69.9%

Mass percentage of O = 30.1%

1.7. 1 mole of CuSO₄ contains 1 mole (1 g atom) of Cu

Molar mass of CuSO₄= 63.5 + 32 + (4 x 16) = 159.5 g mol^-1

Thus, Cu that can be obtained from 159.5 g of CuSO₄ = 63.5 g

∴ Cu that can be obtained from 100 g of CuSO₄ =63.5/159.5 * 100  = 39.81g

1.6. A mass percent of 69% means that 100 g of nitric acid solution contains 69 g of nitric acid by mass.
Molar mass of nitric acid HNO₃= 1 + 14 + (3x16) = 63 gmol^-1