Question

Calculate the modaliti of 2.89 of NaCI dissolved in 0.159 L of water (Given Density of water is 1.00 g/mol)

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Answer 1

[NaCl] = 0.31 m

Step-by-step explanation:

Molality is a sort of solute's concentration, that indicates the moles of solute contained in 1kg of solvent.

In this problem:

solute → NaCl

solvent → H₂O

We need the mass of solvent in kg and we have the volume, so let's convert the volume to mass by the use of density.

1 g/mL = mass / 159 mL

(Notice we needed to convert the volume from L to mL)

159 g is the mass of water. We convert to kg

159 g . 1 kg/1000 g = 0.159 kg

We convert the mass of solute to moles → 2.89 g . 1mol/ 58.45 g = 0.049 moles of NaCl

Molality → mol/kg → 0.049 mol /0.159 kg = 0.31 m