Question

The mass of non volatile solute (molecular weight=40)which should be dissolved in 114gm octane to reduce its vapour pressure is 80%

Answer 1

Answer :

• Mass of non-volatile solute = 10 g.

Step-by-step explanation:

Given that,

• Molar mass of solute, M₂ = 40 g/mol.
• Mass of solvent (octane), w₁ = 114 g.

Also,

• Mar mass of solvent, M₁ = 114 g/mol.

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Suppose $\rm{p^{\circ}}$ be the valour pressure of solvent (octane).

Then, vapour pressure of octane after dissolving the solute will be:

$\implies\rm{\dfrac{80}{100}\times{}p^{\circ}=\dfrac{8}{10}p^{\circ}}$

Assuming that the solution is not dilute, we apply the formula,

$\implies\rm{\dfrac{p^{\circ}-p_s}{p^{\circ}}=\dfrac{w_2/M_2}{w_1/M_1+w_2/M_2}}$

We get (substituting the values),

$\implies\rm{\dfrac{p^{\circ}-\dfrac{8}{10}p^{\circ}}{p^{\circ}}=\dfrac{w_2/40}{114/114+w_2/40}}$

$\implies\rm{\dfrac{2}{10}=\dfrac{w_2/40}{1+w_2/40}}$

$\implies\rm{2=\dfrac{4\times{}w_2}{20}}$

$\implies\rm{w_2=\dfrac{40}{4}}$

$\implies\rm{w_2=}\:\bold{10\:g.}$