Question

What mass in grams of a molecular substance (molar mass = 50.0 g/mol) must be added to 500 g of water to produce a solution that boils at 101.56°c?

Answer 1

M(substance) = 50,0 g/mol.
m(H₂O) = 500 g ÷ 1000 g/kg = 0,5 kg.
Tb(solution) = 101,56°C.
Tb(H₂O) = 100°C.
ΔTb = 101,56°C - 100°C = 1,56°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.
1,56°C = 0,512°C·kg/mol · b.
b = 3,046 mol/kg.
n(substance) = 3,046 mol/kg · 0,5 kg = 1,523 mol.
m(substance) = 1,523 mol · 50 g/mol = 76,15 g.

Answer 2

The mass in grams of the molecular substance to be added is 76.25 grams.

Given the following data:

• Molar mass = 50.0 g/mol

• Mass of water = 500 grams to kg = 0.5 kg

• Boiling temperature of solution = 101.56°C

We know that the temperature at which water boil is 100°C

Molal boiling point constant, Kb (water) = 0.512 °C/m

To find the mass in grams of the molecular substance to be added:

Mathematically, the boiling point elevation of a liquid is given by the formula:

$\Delta T = K_b m$

Where:

• $\Delta T$ is the change in temperature.

• Kb is the molal boiling point constant.

• m is the molality of solution.

Substituting the parameters into the formula, we have;

$101.56 - 100 = 0.512 m\\\\1.56 = 0.512 m\\\\m = \frac{1.56}{0.512}$

m = 3.05 mol/kg

Next, we would determine the number of moles of the molecular substance:

$Number\;of\;moles = m \times mass\;of\;water\\\\Number\;of\;moles = 3.05 \times 0.5$

Number of moles = 1.525 moles

Finally, we solve for the mass of the molecular substance:

$Mass = number\;of\;moles \times molar mass\\\\Mass = 1.525 \times 50$

Mass = 76.25 grams

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