Answer to Question #189856 in General Chemistry for Kristine Olar

Question

A solution of NaCl in 250 grams of water boils at 102. 04 °C (NaCl-58 g/mole) a. What is the concentration of the solution?

b. How many grams of NaCl are present in the solution?

Answer 1

mass of solvent = 250 g = 0.250 kg

molar mass of solute = 58 g/mol

$\text{BP}_{solution} = \text{102.04°C}$

solute = NaCl (Table salt)

solvent = Water

a)

Firstly, calculate the boiling point elevation.

Note: The boiling point of water is 100°C.

$\Delta T_{\text{b}} = \text{BP}_{solution} - \text{BP}_{solvent}$

$\Delta T_{\text{b}} = \text{102.04°C} - \text{100°C}= \text{2.04°C}$

Then, determine the van't Hoff factor.

Since NaCl is an electrolyte, we will assume that NaCl is dissociated completely into one sodium ion and one chloride ion. Therefore, there are two ions in a formula unit of NaCl.

This means that, i = 2

And lastly, calculate the molality of solution.

Note: The value of Kb for water is 0.512°C/molal.

$\text{molality} = \dfrac{\Delta T_{\text{b}}}{iK_{\text{b}}}$

$\text{molality} = \dfrac{\text{2.04°C}}{(2)(\text{0.512°C/molal})}$

$\boxed{\text{molality = 1.99 molal}}$

b)

Firstly, calculate the number of moles of solute.

Note: 1 molal = 1 mol/kg

moles of solute = molality × mass of solvent

moles of solute = 1.9922 mol/kg × 0.250 kg

moles of solute = 0.49805 mol

Then, calculate the mass of solute.

mass of solute = moles of solute × molar mass of solute

mass of solute = 0.49805 mol × 58 g/mol

$\boxed{\text{mass of solute = 28.9 g}}$