Question

0.5 gram of non volatile solute is dissolved in 100 gram of Ethyl Acetate at 20 degree Celsius the vapour pressure of the solution and pure Ethyl Acetate are 72 Torr and 72.8 Torr at-28°c. calculate the molecular weight of the solute.

Answer 1

When a non volatile substance B is added to the solvent,its vapour pressure drops to 0.40 atm ... For dilute solutions containing 2.5g of a non-volatile non electrolyte solute in 100g of  ....

Answer 2

Answer:

The molecular weight of the non-volatile solute, W, measured is $39.82g/mol$.

Explanation:

Data given,

The weight of non-volatile solute = $0.5g$

The weight of ethyl acetate solvent at $20\textdegree C$ = $100g$

The vapour pressure of the solution, p = $72torr$

The vapour pressure of ethyl acetate solvent, p° = $72.8torr$

The molecular weight of the non-volatile solute, W =?

Now, according to Raoult's law:

• $\frac{p\textdegree -p}{p\textdegree} = \frac{n_{solute} }{n_{solute} + n_{solvent} }$   --------equation (1)

Now, the number of moles of solute = $\frac{given weight}{molar weight}$ = $\frac{0.5}{W}$ mol

As we know, the molar mass of ethyl acetate = $88g/mol$

And the number of moles of ethyl acetate = $\frac{given weight}{molar weight}$ = $\frac{100}{88}$ = $1.13mol$

After putting the given values and the calculated values in the equation (1), we get:

• $\frac{72.8 -72}{72.8} = \frac{\frac{0.5}{W} }{\frac{0.5}{W} + 1.13 }$
• $\frac{0.8}{72.8} = \frac{\frac{0.5}{W} }{\frac{0.5 + 1.13 W}{W} }$
• $\frac{0.8}{72.8} = \frac{0.5}{0.5 +1.13W}$
• $0.8(0.5+1.13W) = 72.8\times 0.5$
• $0.4+0.904W = 36.4$
• $0.904W = 36$
• W = $39.82g/mol$

Hence, the molecular weight of the non-volatile solute, W, measured is $39.82g/mol$.