Question

A solution prepared by dissolving 3.27 g of a nonvolatile, covalent solute in 30.0 g of water boils at 100.74 ºC. What is the approximate molar mass of the solute? K bp H 2O = 0.512 ºC/m.

Answer 1

75.2g/mol

Step-by-step explanation:

The normal boiling point of water is 100°C and the boiling point of this solution is 100.74°C.

The elevation in the boiling point (ΔTb) can be calculated using the following equation.

ΔTb = Kb × b

where,

Kb is the ebulloscopic constant (Kb(H₂O) = 0.512°C/m)

b is the molality of the solution

b = ΔTb/Kb

b = (100.74°C-100°C)/(0.512°C/m)

b = 1.45 m

The mass of the solvent is 30.0 g (30.0 × 10⁻³ kg). The moles of solute are:

30.0 × 10⁻³ kg Solvent × (1.45 mol Solute/1 kg Solvent) =0.0435 mol Solute

0.0435 moles of solute have a mass of 3.27 g. The molar mass of the solute is:

M = 3.27g/0.0435 mol = 75.2g/mol

Answer 2

Molar mass of solute is 75.4 g/mol

Step-by-step explanation:

According to Roult's law for a solution of a nonvolatile solute in a volatile solvent-
`\Delta T_(b)=k_(b).m`

where
`\Delta T_(b)` is the ellivation in boiling point of solvent,
`k_(b)` is ebullioscopic constant of solvent and m is molality of solution.

Here solvent is water. Normal boiling point of water is
`100^(0)\textrm{C}`

Hence
`\Delta T_(b)=0.74^(0)\textrm{C}`

If molar mass of solute is M g/mol then 3.27 g of solute =
`(3.27)/(M)` moles of solute.

So molality of solution =

Here
`k_(b)=0.512^(0)\textrm{C}/m`

So,
`0.74=0.512* (((3.27)/(M))* 1000)/(30)`

or, M = 75.4

So molar mass of solute is 75.4 g/mol