Question

97.8 grams of sodium fluoride is put into 350. ml of water.

Calculate the

1. Molarity

2. Molality

3. Mass Percent

4. Mole Fraction

Answer 1

mass of solute( Sodium fluoride = NaF) = 97.8 grams

molar mass of NaF = 41.99 gram/mole

number of moles of NaF = mass of NaF / molar mass = 97.8 / 41.99 = 2.33 moles

number of moles of NaF = 2.33 moles

voluem of water = 350.0 ml = 0.350 L

a)

Molarity = number of moles of the solute / volume of the solution in L

Molarity of sodium fluoride = 2.33 / 0.350 = 6.66 moles/L

Molarity of the sodium fluoride = 6.66 M

b)

volume of the water( Solvent) = 350 mL

density of water = 1.0 g/ml

mass of 1 mL of water = 1.0 grams

mass of 350 mL of water = 1.0 x 350 = 350 grams = 0.350 Kg

Molality = number of moles of the solute / mass of solvent in Kg

Molality of the sodium fluoride = 2.33 / 0.350 = 6.66 moles / Kg

Molality of the sodium fluoride = 6.66 m

c)

mass of solute = 97.8 gramns

mass of solvent = 350 grams

Total mass of the solution = 97.8 + 350 = 447.8 grams

Mass percent = mass of solute * 100 / Total mass of the solution

Mass percent = mass of solute * 100/ mass of solute + mass of solvent

Mass percent = 97.8 * 100 / 447.8

Mass percent = 21.84 %

d)

mass of solute( Sodium fluoride = NaF) = 97.8 grams

molar mass of NaF = 41.99 gram/mole

number of moles of NaF = mass of NaF / molar mass = 97.8 / 41.99 = 2.33 moles

number of moles of NaF = 2.33 moles

mass of water = 350 grams

molar mass of water = 18.1 gram/mole

number of moles of water = 350 / 18.1 = 19.34 moles

Total number of moles = 2.33 + 19.34 = 21.67 moles

Mole fraction of the solute = number of moles of solute / total number of moles of the solution

Mole fraction of the sodium fluoride = 2.33 / 21.67 = 0.107

Mole fraction of the water = number of moles of water / total number of moles the solution

Mole fraction of the water = 19.34 / 21.67 = 0.892