Question

2. A student has a centrifuge tube containing 14.0 g of t-butanol and is asked to make a 1.2 m solution of ethanol/t-butanol. How much ethanol would the student need to add in mL and in g? Show your calculations. Show your calculations. (6 pts)

Answer 1

Answer:

0.774g of ethanol

0.970mL of ethanol

Explanation:

Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.

In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).

14.0g of butanol are 0.014kg and as you want to prepare the 1.2m solution, you need to add:

0.014kg × (1.2moles / kg) = 0.0168 moles of solute = Moles of ethanol

To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:

0.0168moles Ethanol ₓ (46.07g / mol) =

0.774g of ethanol

And to convert mass in g to mL you require density of the substance (Density of ethanol = 0.798g/mL):

0.774g ₓ (1mL / 0.798g) =

0.970mL of ehtanol