What is the molality of a solution containing 100 grams of glucose (C6H12O6, molar mass = 180 g/mol) dissolved in 2.5 kg of water?

0.22 m

72 m

7200 m

40 m

The molality = 0.22 m (option 1 is correct)

Explanation:

Step 1: Data given

Mass of glucose = 100 grams

Molar mass glucose = 180 g/mol

Mass of water = 2.5 kg

Step 2: Calculate moles glucose

Moles glucose = mass glucose / moalr mass glucose

Moles glucose = 100 grams / 180 g/mol

Moles glucose = 0.556 moles

Step 3: Calculate molality

Molality = moles glucose / mass water

Molality = 0.556 moles / 2.5 kg

Molality = 0.2224 molal

The molality = 0.22 m (option 1 is correct)