Question

Calculate the molality of 6g of acetic acid dissolved in 100g benzene

Answer 1

Explanation:

molality = 6×10^-3÷60×10^-3×0.1 =1mol kg-1

Answer 2

Answer:

The molality of $6g$ acetic acid dissolved in $100g$ benzene measured is $1molal$.

Explanation:

Data given,

The mass of acetic acid ( $CH_{3}COOH$ ) ( solute ) = $6g$

The mass of benzene ( solvent ) = $100g$ = $0.1kg$

The molality of the solution =?

As we know,

• Molality = $\frac{Number of moles}{Mass of solvent (kg)}$       -------equation (1)

Firstly, we have to calculate the number of moles of solute.

The molar mass of $CH_{3}COOH$ = $60g/mol$

• Number of moles = $\frac{Given mass}{Molar mass}$ = $\frac{6}{60}$ = $0.1mol$

Now, after putting the values of the number of moles and mass of solvent in equation (1), we get:

• Molality = $\frac{0.1}{0.1}$ = $1molal$

Hence, the molality measured is $1molal$.