Question

Normal boiling point of pure ethyl acetate is 77.06°c. A solution of 50.0 g of naphthalene (c10h8) dissolved in 150 g of ethyl acetate boils at 84.27°c. The boiling point elevation constant of ethyl acetate is

Answer 1

phthalene (C10H8) is dissolved in19.5 g of benzene. Calculate the freezing point of the solution given that the freezing point of pure benzene is 5.5 °C, and the molal freezing point depression constant is 4.45 °C/m.

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Answer 2

Answer:

The boiling point elevation constant of ethyl acetate is $2.77\textdegreee Ckgmol^{-1}$.

Explanation:

Given,

The normal boiling point of ethyl acetate, T = $77.06\textdegree C$

The boiling point of the solution of naphthalene dissolved in ethyl acetate, $T_{b}$ = $84.27\textdegree C$

The naphthalene's mass ( solute ) = $50g$

The mass of ethyl acetate ( solvent ) = $150g$ = $0.15kg$

Ethyl acetate's boiling point elevation constant, $K_{b}$?

As we know,

• We can find out the value of the molal elevation constant:
• $\Delta T = K_{b}\times m$  
• $K_{b} = \frac{\Delta T} {m}$             -------equation (1)

Here,

• ΔT = The change in the boiling point
• m = The molality

Now, firstly, we have to calculate the molality.

As we know,

• The naphthalene's molar mass ($C_{10}H_{8}$)= $128g/mol$.
• The number of moles of naphthalene ( solute ) = $\frac{Given mass}{Molar mass }$ = $\frac{50}{128}$ = $0.39mol$.

Thus,

• The molality = $\frac{Number of moles}{Mass of solvent (kg)}$ = $\frac{0.39}{0.15}$ = $2.6 molal$

Now,

• ΔT = $T_{b} - T$ = $(84.27 - 77.06 )\textdegree C$ = $7.21\textdegree C$

After putting the values of ΔT and molality in equation (1), we get:

• $K_{b} = \frac{7.21} {2.6}$ = $2.77\textdegreee Ckgmol^{-1}$

Hence, the boiling point elevation constant of ethyl acetate is $2.77\textdegreee Ckgmol^{-1}$.