What is the molality of a solution prepared by dissolving 86.9 g of diethyl ether, C4H10O, in 425 g of benzene, C6H6?

The molality of this solution is 2.75 molal

Explanation:

Step 1: Data given

Mass of diethyl ether = 86.9 grams

Mass of benzene = 425 grams = 0.425 kg

Step 2: Calculate moles of solute

Molality = moles of solute/kg of solvent

Molar mass of C4H10O =

4 * molar mass carbon + 10 * molar mass hydrogen + molar mass of oxygen

molar mass C4H10O = 74.12 g/mol

Moles solute (C4H10O) = 86.9 grams / 74.12 g/mol

Moles C4H10O = 1.17 moles

Step 3: Calculate molality

Molality = 1.17 moles of C4H10O / 0.425 kg of C6H6

Molality = 2.75 mol/kg = 2.75 molal

The molality of this solution is 2.75 molal

Molality is 2.76 m/kg

Explanation:

First of all, determinate the moles of solute by dividing mass / molar mass

Molar mass diethyl ether = 74.12 g/m

86.9 g / 74.12 g/m = 1.17 moles

This moles of solute, are in 425 grams of benzene (solvent)

Molality means, the moles of solute which occupy 1kg of solvent.

425 g = 0.425 kg

1.17 m / 0.425 kg = 2.76 m