\(\underset{100}{\mathrm{CaCO}_{\mathrm{g}}} \longrightarrow \mathrm{CaO}_{56 \mathrm{~g}}+\mathrm{CO}_{2}\)
\(\begin{array}{c} \mathrm{MgCO}_{3} \longrightarrow \mathrm{MgO}+\mathrm{CO}_{2} \\ 24+12+48=84 \quad\mathrm{~g} 24+16=40 \mathrm{~g} \end{array}\)
Let the mass of CaCO₃ be 'x', MgCO₃ will be 1 -  x.
100 g of CaCO₃ will gives 56 g of CaO
x g of CaCO₃ will gives \(\frac{56 x}{100} \mathrm{~g} \text { of } \mathrm{CaO}\)
84 g of MgCO₃ gives 40 g of MgO
\( 1-x \mathrm{~g} \text { of } \mathrm{MgCO}_{3} \text { gives }=\frac{40(1-x)}{84} \mathrm{~g} \text { of } \mathrm{MgO} \)
\(\frac{56 x}{100}+\frac{40(1-x)}{84}=0.96 \mathrm{~g} \)
\( 84 \times 0.56 x+40 \times 1.84-40 x=0.96 \times 84 \times 100 \)
\(47.04 x-40 x=80.64-73.60 \)
\(7.04 x=7.04 \Rightarrow x=1 \)
\(\% \text { of } \mathrm{CaCO}_{3}=\frac{1}{1.84} \times 100=54.35 \% \)
\(\% \text { of } \mathrm{MgCO}_{3}=100-54.35=45.65 \% \)

\(2 \mathrm{FeSO}_{4} \stackrel{\text { heat }}{\longrightarrow} \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{SO}_{2}+\mathrm{SO}_{3}\)
Ferrous sulphate, on heating gives Fe₂O₃, SO₂ and SO₃ gases.

\(2 \mathrm{KMnO}_{4}+3 \mathrm{C}_{2} \mathrm{H}_{4}+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{MnO}_{2}+2 \mathrm{KOH}+3\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\)

Hence, the empirical formula in NaCIO₃.

\((i) 4 \mathrm{H}_{3} \mathrm{PO}_{3} \longrightarrow 3 \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{PH}_{3} \)
\((ii) \mathrm{Ca}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{H}_{2}\)
\( (iii)\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+6 \mathrm{NH}_{3}+6 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Fe}(\mathrm{OH})_{3} +3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \)
\((iv) 3 \mathrm{Cl}_{2}+6 \mathrm{NaOH} \stackrel{\text { heat }}{\longrightarrow} 5 \mathrm{NaCl}+\mathrm{NaClO}_{3} +3 \mathrm{H}_{2} \mathrm{O} \)

392 g of sulphuric acid
Molar mass of H₂SO₄ = 2 x 1 + 32 + 4 x 16 = 98 g
98 g of sulphuric acid = 1 mol
392 g of sulphuric acid = 1 mol x\(\frac { 392 \ g }{ (98 \ g) } \) = 4 mol

Molality is defined as the moles of solute per kilogram of solvent.
Molality = m = \(\frac { Moles \ of \ solute }{ Mass \ of \ solvent(in \ kg) } \)
Molality of a solution does not depend on temperature.

(a) 1 mole of N₂ needs 3 moles of H₂ ,
Therefore, 2 moles of N₂ needs 6 moles of H₂ .
But we have only 2 moles of H₂ , so H₂ is limiting reagent.
3 moles of H₂ gives 2 moles of NH₂.
2 moles of H₂ gives \(\frac{2}{3} \times 2=1.33\) moles of NH₃ .
(b) 18 g of water (1 mole) containing 10 x 6.022 x 10²³ electrons = 6.022 x 10²⁴ electrons.
[ \(\because\)1 molecule of H₂O = 2 + 8 = 10 e)
(c) 'M is molarity
'd is density of solution
'm' is molality
\(m=\frac{M \times 1000}{1000 \times d-M \times \text { Molar mass }}\)
\(=\frac{1 \times 1000}{1000 \times 1.10-1 \times 58.5}\)
\(=\frac{1000}{1100-58.5}=\frac{1000}{1041.5}=m=0.96 \mathrm{~mol} / \mathrm{kg} .\)
(d) It is ratio of number of moles of solute (component) to the total number of moles of solute and solvent (all components).
(e) X_glucose + x_H2O = 1
\(\Rightarrow\) X_H2O = 1 - 0.1 = 0.9