If mass percentage of glucose in aqueous solution is 18% then molality of glucose in the solution will be3.5 m1.2 m0.5 m4.1 m

To find the molality of the solution, we need to know the mass of the solvent in kg and the number of moles of solute.

1. First, let's assume we have 100g of the solution. Since the mass percentage of glucose is 18%, this means we have 18g of glucose and 82g of water.

2. Next, we need to convert the mass of glucose to moles. The molar mass of glucose (C6H12O6) is approximately 180g/mol. So, 18g of glucose is 18g / 180g/mol = 0.1 mol.

3. Then, we need to convert the mass of water to kg. Since 1g = 0.001kg, 82g of water is 82g * 0.001kg/g = 0.082 kg.

4. Finally, we can find the molality (m) of the solution by dividing the number of moles of solute by the mass of the solvent in kg. So, m = 0.1 mol / 0.082 kg = 1.22 m.

Therefore, the molality of the glucose in the solution is approximately 1.2 m.