Question

a sodium hydroxide solution that contains 47.8 grams of naoh per l of solution has a density of 1.02 g/ml. calculate the molality of the naoh in this solution.

Answer 1

The molality of the NaOH solution is calculated by dividing the number of moles of NaOH by the mass of the solvent (water) in kilograms, resulting in a molality of 1.229 mol/kg.

Step-by-step explanation:

To calculate the molality of the NaOH solution, we first need to find the number of moles of NaOH. With the given mass of NaOH (47.8 grams) and its molar mass (40.0 g/mol), we can calculate this number.

Moles of NaOH = Mass of NaOH / Molar mass of NaOH

= 47.8 g / 40.0 g/mol

= 1.195 moles

Since the solution's density is 1.02 g/mL, the mass of 1 liter of this solution is:

Mass of solution = Density × Volume

= 1.02 g/mL × 1000 mL

= 1020 grams

To find the molality, we need to know the mass of the solvent (water) in kilograms:

Mass of water = Mass of solution - Mass of solute

= 1020 g - 47.8 g

= 972.2 g

= 0.9722 kg

Now, we can calculate the molality of the NaOH solution:

Molality (m) = Moles of solute / Mass of solvent in kg

= 1.195 moles / 0.9722 kg

= 1.229 mol/kg

Therefore, the molality of the NaOH solution is 1.229 mol/kg.