A solution is prepared by dissolving 25.00 g of sodium chloride (NaCl) in 850.0 g of water. What is the molality of NaCl in this solution?

The molality is 0.504 molal

Explanation:

Step 1: Data given

Mass of NaCl = 25.00 grams

Molar mass NaCl = 58.44 g/mol

Mass of water = 850.0grams

Step 2: Calculate moles NaCl

Moles NaCl = mass NaCl / molar mass NaCl

Moles NaCl = 25.0 grams / 58.44 g/mol

Moles NaCl = 0.428 moles

Step 3: Calculate molality

Molality = moles NaCl / mass H2O

Molality = 0.428 moles / 0.850 kg

Molality = 0.504 molal

The molality is 0.504 molal

0.50m

Explanation:

Data provided in the question:

Mass of the sodium chloride (NaCl) = 25.00 g

Mass of water in the solution = 850.0 g

Now,

We know molality

= Amount of substance (in mol) of solute ] : [ mass (in kg) of the solvent ]

also,

Molar mass of NaCl = 58.44 g/mol

Thus,

Moles of NaCl = 25g : (58.44 g/mol)

= 0.428 mol

Solvent = 850.0 g = 0.85Kg

Thus,

Molality = 0.428 mol : 0.85Kg

= 0.50m