Question

solution of glucose in water is labelled as 10% by mass what would be the molality and molarity of the solution given the density of the solution is 1.2 gram per mole

Answer 1

The molality of the 10% by mass glucose solution is 0.617 mol/kg, and the molarity is approximately 0.666 M, calculated by dividing the number of moles of glucose by the mass and volume of water, respectively.

Step-by-step explanation:

To calculate the molality and molarity of a 10% by mass glucose solution, we first note that a 10% by mass solution means 10 grams of glucose are present in 90 grams of water (since the total mass must be 100 grams to be 10%). To find the molality, we divide the moles of glucose by the kilograms of water. Using the molecular weight of glucose (180.2 g/mol), we find that 10 grams of glucose is approximately 0.0555 moles. Therefore, molality (m) is 0.0555 moles / 0.09 kg = 0.617 mol/kg.

For molarity, we need to convert 100 grams of solution to volume using the given density (1.2 g/mL). This means that 100 grams of solution is equivalent to approximately 83.33 mL (100 g / 1.2 g/mL), or 0.08333 L. Hence, molarity (M) is 0.0555 moles / 0.08333 L = 0.666 M approximately.