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

`m=0.512m`

Step-by-step 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*(1mol)/(60.1g0)=6.656x10^(-3)mol`

Since the molality is computed via:

`m=(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=(6.656x10^(-3)mol)/(0.0130kg)\\ \\m=0.512(mol)/(kg)`

Or just:

`m=0.512m`

Best regards.