A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33° C. Given K_f = 1.86°C/m for water, what is the molar mass of the unknown liquid?
1 Answer
Explanation:
The most important thing to realize here is that since the solute is a liquid, you must be dealing with a covalent compound, which as you know does not ionize in aqueous solution.
This implies that the van't Hoff factor, which tells you the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to
Simply put, every mole of your unknown compound will produce
Your tool of choice here will be the equation for freezing-point depression, which looks like this
color(blue)(DeltaT_f = i * K_f * b)" " , where
The freezing-point depression is defined as
color(blue)(DeltaT_f = T_f^@ - T_f)" " , where
Use the given freezing point of the solution to find
DeltaT_f = 0^@"C" - (-3.33^@"C") = 3.33^@"C"
Now, you strategy here will be to find the molaity of the solution,
So, rearrange the above equation to solve for
DeltaT_f = i * K_f * b " "implies" " b = (DeltaT_f)/(i * K_f)
b = (3.33 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "1.7903 mol kg"^(-1)
Now, a solution's molality is defined as the number of moles of solute divided by the mass of the solvent - expressed in kilograms.
color(blue)(b = n_"solute"/m_"solvent in kg")
Since your solution is said to contain
b = n_"solute"/m_"solvent in kg" " "implies" "n_"solute" = b * m_"solvent in kg"
You will have
n_"solute" = "1.7903 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 90.0 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.16113 moles"
Finally, the molar mass of a substance can be found by dividing the mass of a given sample of that substance by the total number of moles it contains.
color(blue)(M_M = m/n)
In your case, the molar mass of the liquid will be
M_M = "10.0 g"/"0.16113 moles" = "62.062 g mol"^(-1)
Rounded to three sig figs, the answer will be
M_M = color(green)("62.1 g mol"^(-1))