We know mass of 1 mol (6.022 $$ \times $$ 10²³) atoms of carbon = 12 g

If Avogadro number is changed to 6.022 $$ \times $$ 10²⁰ then mass of 1 mol of cabon is

= $${{12 \times 6.022 \times {{10}^{20}}} \over {6.022 \times {{10}^{23}}}}$$ = $$12 \times {10^{ - 3}}\,g$$

$$ \therefore $$ It would change the mass of one mole of carbon.

50 ml of 16.9% solution of AgNO₃

$$\left( {{{16.9} \over {100}} \times 50} \right)$$ = 8.45 g of AgNO₃

n_mole = $${{8.45g} \over {(107.8 + 14 + 16 \times 3)g/mol}}$$

= $$\left( {{{8.45g} \over {169.8g/mol}}} \right) = 0.0497\,moles$$

50ml of 5.8% solution of NaCl contain

NaCl = $$\left( {{{5.8} \over {100}} \times 50} \right) = 2.9g$$

n_NaCl = $${{2.9g} \over {(23 + 35.5)g/mol}}$$

= 0.0495 moles

AgNO3	+	NaCl	$$ \Rightarrow $$	AgCl	+	Na$$ \oplus $$	+	Cl$$ \ominus $$
1 mole		1 mole		1 mole
0.049 mole		0.049 mole		0.049 mole of AgCl

n = $${w \over M}$$

$$ \Rightarrow $$ w = (n_AgCl) $$ \times $$ Molecular mass

= (0.049) $$ \times $$ (107.8 + 35.5) = 7.02 g

n_Mg = $$1 \over 24$$ = 0.0416 moles

n_O2 = $$0.56 \over 32$$ = 0.0175 moles

	Mg	+	$${1\over2} O_2$$	$$ \Rightarrow $$	MgO
Initial	0.0416 moles		0.0175 moles		0
Final	( 0.0416 - 2 x 0.0175) = 0.0066 moles		0		2 x 0.0175

$$ \because $$ Mass of Mg left in excess = 0.0066 $$ \times $$ 24 = 0.16 g

According to Avogadro's hypothesis ratio of the volumes of gases will be equal to the ratio of their no. of moles.

So, no of moles = $${{Mass} \over {Mol.\,\,mass}}$$

n_H2 = $${w \over 2}$$; n_O2 = $${w \over 32}$$; n_CH4 = $${w \over 16}$$

So, the ratio will be $${w \over 2}$$ : $${w \over 32}$$ : $${w \over 16}$$ or 16 : 1 : 2