Given:
mass of solvent = 250 g = 0.250 kg
molar mass of solute = 58 g/mol
[tex]\text{BP}_{solution} = \text{102.04°C}[/tex]
solute: NaCl
solvent: water
Required:
a) molality
b) mass of solute
Solution: (a)
Step 1: Calculate the boiling point elevation.
Note: The boiling point of water is 100°C.
[tex]\Delta T_{\text{b}} = \text{BP}_{solution} - \text{BP}_{solvent}[/tex]
[tex]\Delta T_{\text{b}} = \text{102.04°C} - \text{100°C}[/tex]
[tex]\Delta T_{\text{b}} = \text{2.04°C}[/tex]
Step 2: 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,
[tex]i = 2[/tex]
Step 3: Calculate the molality of solution.
Note: The value of Kb for water is 0.512°C/molal.
[tex]\text{molality} = \frac{\Delta T_{\text{b}}}{iK_{\text{b}}}[/tex]
[tex]\text{molality} = \frac{\text{2.04°C}}{(2)(\text{0.512°C/molal})}[/tex]
molality = 1.9922 molal
[tex]\boxed{\text{molality = 1.99 molal}}[/tex]
Solution: (b)
Step 1: 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
Step 2: Calculate the mass of solute.
mass of solute = moles of solute × molar mass of solute
mass of solute = 0.49805 mol × 58 g/mol
[tex]\boxed{\text{mass of solute = 28.9 g}}[/tex]
[tex]\\[/tex]
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