[Solved] According to Raoult's law the relative lowering of (2024)

Explanation:

Raoult's law is stated as:

The relative lowering in vapor pressure of an ideal solution containing the non-volatile solute is equal to the mole fraction of the solute at a given temperature.

i,e, Relative lowering of v.p = mole fraction of solute

Mathematically, it is given by:

\(\frac{{P_1^0 - {P_1}}}{{P_1^0}} = {x_2} = \frac{{{n_2}}}{{{n_1} + {n_2}}}\)

Where n₁and n₂= no. of moles of solvent and solute respectively, x₂= mole fraction of solute, P₁^o- P₁= reduction in vapor pressure of the solvent.

Key Points

• \(Mole\;fraction\;X:\;moles\;of\;a\;component\;/\;Total\;moles\;of\;solution\)
• Mass %: Mass of solute (in g) present in 100g of solution.
• Mass/Vol: Mass of solute (in g) present in 100 mL of solution.
• v/v: Volume of solute/volume of solution {only for liq-liq solutions.}
• g/L: Wt. of solute (g) in 1 L of solutions.
• \(Molality\;\left( m \right) = \frac{{mass\;of\;solute}}{{mass\;of\;solvent\;\left( {kg} \right)}}\)
• \(Molarity\;\left( M \right) = \frac{{moles\;of\;solute}}{{volume\;of\;solution\left( L \right)}}\)
• \(ppm = \frac{{moles\;of\;solute}}{{mss\;of\;solution}} \times {10^6}\)