What is the molality of a glucose solution prepared by dissolving 18.0g of glucose, C6H12O6 in 125g of water?
A)0.000794
B)0.144
C)0.699
D)0.794
Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water.
Example:
What is the molality of a solution of 10 g NaOH in 500 g water?
Solution:
10 g NaOH / (4 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.05 M / kg
molality = 0.50 m
hope this helps.....let me know if you need more help
The example looks good but there is a typo. 4g NaOH should read 40 g NaOH/mol.
WHAT IS 1MOLAL SOLUTION
Thank you for pointing out the typo. You are correct that the molar mass of NaOH is 40 g/mol. Therefore, the correct calculation would be:
10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.50 M / kg
I apologize for the mistake and thank you for bringing it to my attention. Is there anything else I can assist you with?
The correct example should be:
10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.5 M/kg
Now let's calculate the molality of the glucose solution:
18.0 g glucose / (180.18 g glucose/mol) = 0.1 mol glucose
125 g water x 1 kg / 1000 g = 0.125 kg water
molality = 0.1 mol / 0.125 kg
molality = 0.8 M/kg
Therefore, the molality of the glucose solution is 0.8 M/kg. The closest answer choice is (D) 0.794, so the correct answer is (D) 0.794.