किसी पदार्थ के \( 7.2 g \) को \( 100 g \)

1.	किसी पदार्थ के \( 7.2 g \) को \( 100 g \) पानी में घोलने से वाष्पदाब का आपेक्षिक अवनमन \( 0.00715 \) होता है । पदार्थ का अणु द्रव्यमान ज्ञात कीजिए।
Answer» Let say, molecular weight of substance is x g/mol. As we know, relative lowering of vapour pressure is equal to the mole fraction of solute. ∴ \(\frac{P^o-P_s}{P_o}=x_2\)------(1) where, \(\frac{P^o-P_s}{P^o}\) = relative lowering in vapour pressure = 0.00715 x₂ = mole fraction of substance (solute) Number of moles of water = 100/18 = 5.56 mol Number of moles of substance = 7.2/x = mol ∴ x₂ = \(\cfrac{\frac{7.2}x}{5.56+\frac{7.2}x}\) ∵ the moles of solvents are always much greater than moles of solution. So that we can neglect 7.2/x as compare to 5.56. ∴ x₂ = \(\cfrac{\frac{7.2}x}{5.56}\) ⇒ x = \(\frac{7.2}{5.56\times0.00715}\) x = 181.1 g/mol Hence, the molecular weight of substance will be 181.1g/mol.

किसी पदार्थ के \( 7.2 g \) को \( 100 g \) पानी में घोलने से वाष्पदाब का आपेक्षिक अवनमन \( 0.00715 \) होता है । पदार्थ का अणु द्रव्यमान ज्ञात कीजिए।

Answer»

Let say, molecular weight of substance is x g/mol.

As we know, relative lowering of vapour pressure is equal to the mole fraction of solute.

∴ \(\frac{P^o-P_s}{P_o}=x_2\)------(1)

where, \(\frac{P^o-P_s}{P^o}\) = relative lowering in vapour pressure = 0.00715

x₂ = mole fraction of substance (solute)

Number of moles of water = 100/18 = 5.56 mol

Number of moles of substance = 7.2/x = mol

∴ x₂ = \(\cfrac{\frac{7.2}x}{5.56+\frac{7.2}x}\)

∵ the moles of solvents are always much greater than moles of solution. So that we can neglect 7.2/x as compare to 5.56.

∴ x₂ = \(\cfrac{\frac{7.2}x}{5.56}\)

⇒ x = \(\frac{7.2}{5.56\times0.00715}\)

x = 181.1 g/mol

Hence, the molecular weight of substance will be 181.1g/mol.