Question

A solution prepared by dissolving 180.0 mg of a sugar (a molecular compound and a nonelectrolyte) in 1.00 g of water froze at -1.86°C. What is the molar mass of this sugar? The value of Kf is 1.86°C/m._______ g/mol

Answer 1

180 g/mol.

Step-by-step explanation:

What is given?

ΔT = [0 - (- 1.86 )] °C = 1.86 °C.

i = 1 (for a nonelectrolyte).

Kf = 1.86 °C/m.

Step-by-step solution:

To solve this problem, we have to use the boiling point elevation formula, which is the following:

`\Delta T=i\cdot m\cdot K_f.`

Where ΔT is the change in boiling point, i is the Van't Hoff factor, m is the molality of solution, and Kf is the molal boiling point constant.

Let's calculate the molality with the given data:

`m=(\Delta T)/(i\cdot K_f)=\frac{1.86\text{ \degree C}}{1\cdot1.86\text{ }(\degree C)/(m)}=1\text{ m.}`

1 m is the same that 1 mol/kg. As we have 1.00 g of water and 180.0 mg of the sugar, we can multiply 1 mol/kg by the mass of water. Remember that 1 kg equals 1000 g, so 1.00 g is the same that 0.001 kg:

`1\text{ }(mol)/(kg)\cdot0.001\text{ kg=0.001 mol.}`

Remember that the units of the molar mass is in g/mol, and 1 g equals 1000 mg, so 180.0 mg is the same that 0.18 g. If we divide 0.18 g by 0.001 mol, we will obtain the molar mass of the sugar, which would be:

`\begin{gathered} Molar\text{ mass=}\frac{0.18\text{ g}}{0.001\text{ mol}}, \\ Molar\text{ mass=180}(g)/(mol). \end{gathered}`

The answer would be that the molar mass is 180 g/mol.