# Question 19ef4

May 20, 2015

$\text{4.2 g}$

#### Explanation:

You have $\text{4.2 g}$ of ethane in that much hexane solution.

Percent concentration by mass is defined as the mass of solute, in your case ethane, divided by the total mass of the solution, and multiplied by 100%.

In order to be able to calculate the amount of ethane present in that much $\text{13.9% m/m}$ hexane solution, you need to first figure out what the actual weight of the solution is.

To do that, use its density and volume

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

${m}_{\text{solution" = 0.611"g"/cancel("mL") * 50cancel("mL") = "30.55 g}}$

You know that the mass of the solute accounts for $\text{13.9%}$ of the solution, which means that

"13.9%" = m_"ethane"/m_"solution" * 100% => m_"ethane" = ("13.9%" * m_"solution")/(100%)

${m}_{\text{ethane" = (13.9cancel(%) * 30.55)/(100cancel(%)) = "4.246 g}}$

Now, I'm going to give the answer rounded to two sig figs, even though you only gave one sig fig for the volume of the solution

m_"ethane" = color(green)("4.2 g")

May 20, 2015

4.2g

${M}_{\text{eth")/M_("soln}} \times 100 = 13.9$

So:

M_("eth")=(13.9xxM_("soln"))/(100)

M_("soln")="density"xx"volume"=0.611xx50=30.55"g"

So:

M_("eth")=(13.9xx30.55)/(100)=4.2"g"#