Question

What is the molecular mass for a non-electrolyte if 35.0 g of it is dissolved in 45.0 grams of water and the solutions boiling point is 101.25oC? (The KB for H2O is .51°C/m)

Answer 1

The boiling point of water increases as the amount of impurities dissolved in it increases. For our purposes, we will consider the non-electrolyte to be the dissolved impurity. The change in the boiling point can be calculated using the equation:

$\Delta T_b = i \times K_b \times m$

where $\Delta T_b$ is the change in boiling point, $i$ is the van ‘t Hoff factor (whose value denotes the number of particles each formula unit of the dissolved substance dissociates into in water), $K_b$ is the boiling point elevation constant, and $m$ is the molality (moles of solute/kilogram of solvent) of the solution.

Right off the bat, since we're dealing with a non-electrolyte, the dissolved substance can be assumed not to dissociate in water. So, our van ‘t Hoff factor, $i$, would be 1 (by contrast, the $i$ for an ionic compound like NaCl would be 2 since, in water, NaCl would dissociate into two particles: one Na⁺ ion and one Cl⁻ ion). We're also given our $K_b$, which is 0.51 °C/m.

Assuming the normal boiling point of pure water to be 100 °C (a defined value for sig fig purposes), the change in boiling point from having dissolved 35.0 g of the non-electrolyte can be obtained by subtracting 100 °C from the final—elevated—boiling point of 101.25 °C:

$\Delta T_b = 101.25\text{ }^o\text{C} - 100\text{ }^o\text{C} = 1.25\text{ }^o\text{C}$

Now, recall what we're asked to determine: the molecular mass of the dissolved substance. There is one unknown left in the equation: the molality of the solution. Let's first solve for that:

$m = \frac{\Delta T_b}{K_b} = \frac{1.25^\text{ o}\text{C}}{0.51^\text{ o}\text{C}/m} \\ m = 2.45 \text{ mol solute/kg water}.$

Notice that we didn't include the i since its value is 1.

Now, what would happen if we multiplied our molality by the mass of water we've been given? We would be left with the moles of solute. And what are we asked to find? The molecular mass, or the mass per mole. We can accomplish this in two steps. Remember to convert your mass of water to kilograms:

$2.45 \text{ mol solute/kg water} \times 0.045 \text{ kg water} = 0.110 \text{ mol solute.}$

And, finally, we divide the mass of our solute by the number of moles of solute:

$\frac{35.0 \text{ g solute}}{0.110 \text{ mol solute}} = 317.5 \text{ g/mol}$

Our answer to two significant figures (which is the number of sig figs to which our $K_b$ is given) would be 320 g/mol.