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