Emphirical formula mass = CH₂O
= 12 + 2\(\times\)1 + 16
= 30 
Moleular mass = 180 
n = \( \frac{Molecular\ mass}{Emphirical\ formula\ mass}\)
= \(\frac{180}{30}\)
= 6
\(\therefore\)  Molecular formula = n\(\times\) Emphirical formula 
= 6\(\times\)CH₂O
= C₆H₁₂O₆

Law of conservation of mass and Avogadro's law because law of conservation of mass is simply the law of indestructibility of matter during physical or chemical changes.
Avogadro law states that equal volumes of different gases contain the same number of molecules under similar conditions of temperature and pressure.

One mole of any substance contains 6.022\( \times\)10²³ atoms /molecules
Hence, number of millimoles of H₂SO₄
= molarity \( \times\) volume in mL
= 0.02 \( \times\)100
= 2millimoles
=  2\( \times\) 10^-3 mol 
Number of molecules = number of moles \( \times\) N_A  
= 2\( \times\)10^-3\( \times\)6.022\( \times\)10²³
=12.044 x 10²⁰ molecules

Given that, density of solution = 3.12 g mL^-1
Volume of solution = 1.5 mL
For a solution, Mass = Volume x density
= 1.5 mL x 3.12 g mL^-1 = 4.68 g
The digit 1.5 has only two significant figures, so the answer must also be limited to two significant figures. So, it is rounded off to reduce the number of significant figures. Hence, the answer is reported as 4.7 g.

For the reaction
\({ H }_{ 2 }{ SO }_{ 4 }+2NaOH{ \longrightarrow }{ Na }_{ 2 }{ SO }_{ 4 }+H_{ 2 }O\) 
1L: of 0.1 M H₂SO₄ contains = 0.1 mole of H₂SO₄
1L of 0.1 M Na OH contains = 0.1 mole of NaOH
According to the reaction,1 mole of  H₂SO₄ reacts with 2 moles of NaOH. Hence,0.1 mole of NaOH will react with 0.05 mole of H₂SO₄ 
(and 0.05 mole of H₂SO₄ will be left unreacted), i.e NaOH is the limiting reactant. Since, 2 moles of NaOH produces 1 mole of Na₂SO₄.
Hence,0.1 mole of NaOH will produces 0.05 mole of Na₂SO₄ 
Mass of Na₂SO₄ = moles \(\times \)molar mass
= 0.5\(\times \) (46 + 32  + 64)g = 7.10g
Volume of solution after mixing = 2L
Since only 0.05 mole of H₂SO₄ is left behind. as NaOH is completely used in the reaction.
Therefore,molarity of the given solution is calulated from moles of H₂SO₄.
H₂SO₄ left unreacted in the solution = 0.05 mole
\(\therefore \) Molarity of the solution \(=\frac { 0.05 }{ 2 } =0.025\)mol L^-1