According to Dalton's atomic theory, an atom is the ultimate particle of matter which cannot be further divided.

SI unit of molarity = mol dm^-3

(a) Steps:
108   Ã⋅7.2 = 14.5833
Three S.F. Two S.F.
Here answer should have two significant figures. Therefore, correct answer, after rounding off two significant figures is 15.
(b) Steps: 2 x 1.6 x 10²
2 x 163.2 = 326.4 = 300
(c) Steps: (1.0042 - 0.0034) x 1.23
1.0008 x 1.23 = 1.230984
= 1.23

N₂O₅,CCL₄,C₆,H₁₂O₆,C₂H₆O.

A = 14g

14 g

0.33 mol

(a) 5
(b) 5
(c) 1
(d) 4

(a) 91 x 10^-25
(b) 0.7 kg

1.54 ×10³pm μ s ^-1.

32.39

6.509

Density = \(\frac { Mass }{ Volume({ m }^{ 3 }) }\) 
Mass = 57.642 g = 57.642 x 10^-3 kg
Volume = 49.0 cm³ = 49.0 x (10^-2)³ = 49.0 x 10^-6m³
\(\therefore \)  Density = \(\frac { 57.642\times { 10 }^{ -3 } }{ 49.0\times { 10 }^{ -6 } } = 1.176 x 103 kg/m3\)

Vapour density of the mixture of NO₂ and N₂O₄ = 38.3
Molecular mass of the mixture = 2 x Vapour density = 2 x 38.3 = 76.6 u = 76.6 g
Mass of the mixture = 100 g
No. of moles of the mixture = \(\frac { 100 }{ 76.6 } \)
Let the mass of NO₂ in the mixture = x g
\(\therefore \) Mass of N₂O₄ in the mixture = (100 - x) g
Molar mass of NO₂ = 14 + 32 = 46 u = 46 g
Molar mass of N₂O₄ = 28 + 64 = 92 u = 92 g
No. of moles of NO2 = \(\frac { x }{ 46 } \)
No. of moles of N₂O₄ = \(\frac { (100-x) }{ 92 } \)
Total no. of moles in the mixture = \(\frac { x }{ 46 } \)+\(\frac { (100-x) }{ 92 } \)
Equating (i) and (ii),\(\frac { x }{ 46 } \)+\(\frac { (100-x) }{ 92 } \) = \(\frac { 100 }{ 76.6 } \)
92x + 46(100 - x) = \(\frac { 100 }{ 76.6 } \)x 46 x 92 = 5524.8
92x - 46x = 5524.8 - 4600 = 924.8.