Some Basic Concepts of Chemistry

	Element	%	Atomic mass	Relative no. of atoms	Simplest ratio of atoms
1	C	40	12	${{40} \over {12}}$ = 3.33	${{3.33} \over {3.3}}$ = 1
2	H	13.3	1	${{13.3} \over 1}$ = 13.3	${{13.3} \over {3.3}}$ = 4
2	N	46.7	14	${{46.7} \over {14}}$ = 3.3	${{3.3} \over {3.3}}$ = 1

$ \therefore $ The empirical formula is CH₄N.

1 mole = 6.023 $ \times $ 10²³ number of molecules.

1 g mole of O₂ = 32 g of O₂

$ \Rightarrow $ 16 g of O₂ = 0.5 g mole of O₂

1 g mole of N₂ = 28 g of N₂

$ \Rightarrow $ 7 g of N₂ = 0.25 g mole of N₂

1 g mole of H₂ = 2 g of H₂

$ \Rightarrow $ 2 g of H₂ = 1 g mole of H₂

1 g of NO₂ = 14 + 16 $ \times $ 2 = 46

$ \Rightarrow $ 16 g of NO₂ = 0.35 mole NO₂

Molarity = ${{1.17 \times 1000} \over {36.5 \times 1}} = 32.05$

${{100} \over {0.5}} \times 78.4 = 1.568 \times {10^4}$

Specific volume ( vol. of 1 g) cylindrical virus particle

= 6.02 $ \times $ 10^-2 cc/g

Radius of virus, r = 7 Å = 7 $ \times $ 10^-8 cm

Volume of virus = $\pi {r^2}l$

= ${{22} \over 7} \times (7 \times {10^{ - 8}}) \times 10 \times {10^{ - 8}}$

= 154 $ \times $ 10^-23 cc

wt. of one virus particle = ${{Volume} \over {Specific\,\,Volume}}$

$ \Rightarrow $ ${{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}}g$

$ \therefore $ Molecular wt of virus = wt. of N_A particle

= ${{1.54 \times {{10}^{ - 23}}} \over {6.02 \times {{10}^{ - 2}}}} \times 6.02 \times 10^{-23} \,\,g/mol$

= 15400 g/mol = 15.4 kg/mol

BaCO₃ $ \to $ BaO + CO₂

197.34 g $ \to $ 22.4 L at N.T.P

9.85 g $ \to $ ${{22.4} \over {197.34}} \times 9.85 = 1.118\,L$

In A₃(BC₄)₂, (+2) $ \times $ 3 + 2[+5+4(-2)]

$ \Rightarrow $ +6 + 10 -16 = 0

$ \therefore $ Oxidation numbers of A, B, C are + 2, +5 and $-$2 respectively.