Question

Calculate the molarity and molality of 49.0 wt% HF, assuming a density of 1.16 g/mL.

Answer 1

The molarity of the 49.0 wt% HF solution is 2.449 M, and the molality is 24.49 mol/kg.

Step-by-step explanation:

To calculate the molarity and molality of a 49.0 wt% HF solution, we need to know the molar mass of HF, which is 20.01 g/mol. First, we need to convert the weight percent to grams. If we have 100 g of the solution, 49.0 g would be HF. Next, we need to calculate the number of moles of HF by dividing the mass by the molar mass: 49.0 g / 20.01 g/mol = 2.449 mol. The molarity is then calculated by dividing the number of moles by the volume (in L) of the solution. If we assume a volume of 1 L, the molarity would be 2.449 M.

To calculate the molality, we need to know the mass of the solvent (water in this case) and its density. Let's assume a mass of 100 g for water. The molality is then calculated by dividing the number of moles of solute by the mass of the solvent (in kg): 2.449 mol / 0.1 kg = 24.49 mol/kg.

Answer 2

Answer:
Molarity 28.42
Molality = 24.5

To answer this question, it would be easier if you assume that the liquid volume is 1L. You also need to know that HF weight is 20 (fluor=19, helium=1). Then you need to calculate how many moles of HF inside the solution. The calculation would be:

1L x (1000ml/L) x 1.16g/ml x (49g/100g) / (20g/mole)= 28.42 mole

Then the solution molarity: 28.42mole / 1L= 28.42
Then the solution molality: 28.42mole / (1L x 1000ml/L x 1.16g/ml) = 24.5