Question

What is the molar mass of a non-electrolyte compound if 4.28 grams is dissolved in 25.0 grams of chloroform solvent to form a solution which has a boiling point elevation of 2.30 degrees celsius?

The boiling constant of chloroform is 3.63 degrees Celsius/m
im not sure if this is needed but molar mass of chloroform = 119.38 g/mol

I need a step by step explanation please.

Answer 1

The molar mass is 270 g/mol.

Explanation:

The formula for boiling point elevation `ΔT_"b"` is

`color(blue)(bar(ul(|color(white)(a/a)ΔT_"b" = K_"b"bcolor(white)(a/a)|)))" "`

where

`K_"b"` = the molal boiling point elevation constant of the solvent
`bcolor(white)(m)` = the molality of the solution

We can rearrange the formula to get

`b = (ΔT_"b")/K_"b"`

In your problem,

`ΔT_"b" = "2.30 °C"`
`K_"b" = "3.63 °C·kg·mol"^"-1"`

∴ `b = (2.30 color(red)(cancel(color(black)("°C"))))/(3.63 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "0.6336 mol·kg"^"-1"`

You have 4.28 g of compound in 0.0250 kg solvent.

;: `b = "0.6336 mol"/(1 color(red)(cancel(color(black)("kg")))) = "4.28 g"/("0.0250" color(red)(cancel(color(black)("kg"))))`

Divide both sides of the equation by `0.6336`.

∴ `"1 mol" = "4.28 g"/0.0250 × 1/0.6336 = "270 g"`

The molar mass is 270 g/mol.