Question

3. A student adds 0.400g of n-propanol to 13.0 g of t-butanol. What is the molality of the solution? Show your calculations. (3 pts

Answer 1

Answer:

$m=0.512m$

Explanation:

Hello,

In this case, we can consider the n-propanol as the solute (lower amount) and the t-butanol as the solvent (higher amount), for which, initially, we must compute the moles of n-propanol (molar mass = 60.1 g/mol) as shown below:

$n_{solute}=0.400g*\frac{1mol}{60.1g0}=6.656x10^{-3}mol$

Since the molality is computed via:

$m=\frac{n_{solute}}{m_{solvent}}$

Whereas the mass of the solvent is used in kilograms (0.0130g for the given one). Thus, we compute the resulting molality of the solution:

$m=\frac{6.656x10^{-3}mol}{0.0130kg}\\ \\m=0.512\frac{mol}{kg}$

Or just:

$m=0.512m$

Best regards.