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A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Calculate the density, mole fraction, molarity, and molality of the solution.

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Answer:

Density is: 1.05 g/ml

Mole fraction solute: 0.015

Mole fraction solvent: 0.095

Molarity: 0.80 M

Molality: 0.82 m

Step-by-step explanation:

A typical excersise of solution.

It is more confortable to make a table for this.

| masss | volume | mol

solute | | |

solvent | | |

solution | | |

Let's complete, what we have.

| masss | volume | mol

solute | 10.8g | |

solvent | | 133 mL |

solution | | 137 mL |

We can first, know how many moles are 10.8 g

Molar Mass H3PO4 = 97.99 g/mol

Mass / Molar mass = mol

10.8 g / 97.99 g/m = 0.110 mol

Density of water is 1 g/ml (it is a very knowly value)

From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

| masss | volume | mol

solute | 10.8g | | 0.110 mol

solvent | 133g | 133 mL | 7.39 mol

solution | 143.8g | 137 mL | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles or solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

User Sergey Gavruk
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