Question

Calculate the weight of non- volatile solute having molecular weight $40$ , which should be dissolved in $57g$ octane to reduce its vapour pressure to $80\%$
A.$47.2g$
B.$4g$
C.$106.2g$
D.None of these

Answer 1

If the solute has the tendency to evaporate itself, that type of solute is known as non- volatile solute. And it does not make a vapour pressure in that solution. So it can sassily escape from the solution in the form of gas. But in the case of volatile solute, that solute should produce the vapour pressure at the time of boiling the solution.

Complete answer:
The weight of non- non-volatile solute is not equal to$47.2g$. Hence, option (A) is incorrect.
According to the question, given,
The molecular weight of non- volatile solute is equal to 40 and the given weight of octane is$57g$. And this octane will reduces its vapour pressure to $80\%$
Consider the vapour pressure of pure octane is equal to $p_1^0$.
Then, after dissolving the non- volatile solute, the vapour pressure of octane$= \dfrac{{80}}{{100}}$
The vapour pressure of pure octane is equal to the vapour pressure of octane after dissolving the non- volatile solute.
Hence, $p_1^0 = 0.8p_1^0$
Molar mass of solute, ${M_1} = 40g$
Molar mass of octane, $\left( {{C_8}{H_{18}}} \right),{M_1} = 114g/mol$
Let’s see the relation,
$\dfrac{{p_1^0 - {p_1}}}{{p_1^0}} = \dfrac{{{w_2} \times {M_1}}}{{{M_2} \times {w_1}}}$
By substituting the values in above equation,
$\dfrac{{p_1^0 - 0.8p_1^0}}{{p_1^0}} = \dfrac{{{w_2} \times 114}}{{40 \times 57}}$
Rearrange and simplifying the equation will get the weight of non- volatile solute,
$\dfrac{{0.2p_1^0}}{{p_1^0}} = \dfrac{{{w_2}}}{{20}}$

0.2 = \dfrac{{{w_2}}}{{20}} \\\ {w_2} = 4g \\\ \end{gathered} $$ Hence, option (B) is correct. If the non- volatile solute is dissolved in $$57g$$ octane to reduce its vapour pressure to $$80\% $$ then the weight of non – volatile solute is not equal to $$106.2g$$. Hence, option (C) is incorrect. The weight of non- non-volatile solute is equal to $$4g$$. Hence, the option (D) is incorrect. **Note:** Here, the non- volatile solvent is dissolved in the solvent octane to reduce its vapour pressure to$$80\% $$. The vapour pressure is also known as equilibrium vapour pressure and it is the leaning of a substance to change its phase into vapour phase or gaseous phase. The vapour pressure of non- volatile solute is equal to the vapour pressure of the pure solvent at which the product of mole fraction and temperature.