Question

Determine the molality of a solution of benzene dissolved in toluene (methylbenzene) for which the mole fraction of benzene is 0.176. Give your answer to 2 decimal places

Answer 1

The molality of the solution of benzene dissolved in toluene is 0.00543 mol/kg.

Step-by-step explanation:

The molality of a solution can be calculated using the mole fraction and the molar mass of benzene and toluene. The mole fraction of benzene can be calculated by dividing the moles of benzene by the total moles of benzene and toluene. The moles of benzene can be determined using the given volumes of benzene and toluene and their respective densities. The molar mass of benzene is 78.11 g/mol. By substituting the values into the formula, we can calculate the molality of the solution to be 0.176 * 78.11 g/mol / (1000 g + 0.867 g/mL * 100 mL + 0.874 g/mL * 300 mL) = 0.00543 mol/kg.

Answer 2

2.32 m

Step-by-step explanation:

So, according to definition of mole fraction:

`Mole\ fraction\ of\ benzene=\frac {n_(benzene)}{n_(benzene)+n_(toluene)}`

Mole fraction = 0.176

Applying values as:

`0.176=\frac {n_(benzene)}{n_(benzene)+n_(toluene)}`

`0.176* ({n_(benzene)+n_(toluene)})={n_(benzene)}`

So,

`0.176* n_(toluene)}=0.824* {n_(benzene)}`

`{n_(benzene)}=\frac {0.176}{0.824}* n_(toluene)}`

`{n_(benzene)}=0.2136* n_(toluene)}`

Also, Molar mass of toluene = 92.14 g/mol

Thus,

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

`Mass=92.14* n_(toluene)}\ g`

Also, 1 g = 0.001 kg

So,

`Mass\ of\ toluene=0.09214* n_(toluene)}\ kg`

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,

`Molality\ (m)=\frac {0.2136* n_(toluene)}{0.09214* n_(toluene)}`

Molality of benzene = 2.32 m