Calculate the molality of a solution containing 257 g glucose (C6H12O6) dissolved in 1.62 L of water. (Assume a density of 1.00 g/mL for water.)

Solution 100P :

Step 1:

Here, we have to calculate the molality of each solution :

First, let’s see what is molality :

Molality(m) is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is mol/kg.

Formula to find the molality of solution :

Molality(m) =

Step 2:

Given :

Mass of solute - glucose (C6H12O6) = 257 g

Mass of solvent - water (H2O) = 1.62 L

(Assume a density of 1.00 g/mL for water.)

We have solute in grams, we need moles of solute to calculate the molality. Hence, first we find the moles of solute from the grams given :

Number of moles(n) =

Molar mass of glucose (C6H12O6) = 180.15 g/mol.

Therefore, n =

n = 1.426 mol

We need mass of solvent in Kg, to calculate molality, hence let’s convert mass of water in L to Kg :

1 L = 1 Kg

Therefore, 1.62 L of water in Kg will be 1.62 Kg