(a) 1 gram atom of Mg = 24g
2.5 gram atoms of Mg = 24 x 2.5 = 60g
(b) 1 gram atom of N = 14g;
14g of N = 1 gram atom
1.4g of N = \(\frac { 1 }{ 14} \)x ​​​​​​1.4 = 0.1 gram atom.

8g of calcium
Gram atomic mass of Ca = 40 g
40 g of calcium = 1 mol
8.0 g of calcium = 1 mol x\(\frac { (8.0 \ g) }{ (40 \ g) } \)=0.2 mol

44.8 litres of sulphur dioxide at N. T.P.
22.4 litres of sulphur dioxide at N.T.P. = 1 mol
44.8 litres of sulphur dioxide at N.T.P. =\(\frac { 1 \ mol }{ (22.4 \ L) } \) x (44.8L) = 2.0 mol

2.000 x 10³

8.721 x 10⁴

6.509

(a) 76.47%
(b) 68.42%
(c) 52.00%

(i) \(4 \times 10^{19} \mathrm{~m}(\mathrm{ii}) 1.4 \times 10^9 \mathrm{~m}(\mathrm{iii}) 41 \times 10^{15} \mathrm{~m}\)
(i) \(1 \mathrm{Em} \text{,i.e., exametre }=18^{18} \mathrm{~m} \therefore 40 \mathrm{Em}=40 \times 10^{18} \mathrm{~m}=4 \times 10^{19} \mathrm{~m}\).
(ii) \(1 \mathrm{Gm}\text{, i.e., gigametre }=10^9 \mathrm{~m} \therefore 1.4 \mathrm{Gm}=1.4 \times 10^9 \mathrm{~m}\)
(iii) \(1 \mathrm{Pm}\text{, i.e., petametre }=10^{15} \mathrm{~m} \therefore 41 \mathrm{Pm}=41 \times 10^{15} \mathrm{~m}\)

Step 1. Conversion of mass per cent to grams : Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.
Step 2. Convert into number moles of each element : Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound
Moles of hydrogen = \(\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g}}=4.04\)
Moles of carbon = \(\frac{24.27 \mathrm{~g}}{12.01 \mathrm{~g}}=2.021\)
Moles of chlorine = \(\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g}}=2.021\)
Step 3. Divide each of the mole values obtained above by the smallest number amongst them : Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl. In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.
Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements : CH₂Cl is, thus, the empirical formula of the above compound.
Step 5. Writing molecular formula : (a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass is
12.01 + (2 x 1.008) + 35.453
= 49.48 g
(b) Divide Molar mass by empirical formula mass
\(\frac{\text { Molar mass }}{\text { Empirical formula mass }}=\frac{98.96 \mathrm{~g}}{49.48 \mathrm{~g}}\)
= 2 = (n)
(c) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = CH₂Cl, n = 2. Hence molecular formula is C₂H₄Cl₂.

= 0.364:99.636