Question #aaf98
1 Answer
Here's what I got.
Explanation:
From what I can tell by looking at the problem, you don't really need to know the density of water.
Your strategy here will be to pick a sample of this
- the mass of the solute, which in your case is sodium hydroxide
- the mass of the solvent, which in your case is water
So, a solution's percent concentration by mass is defined as the mass of the solute divided by the total mass of the solution, and multiplied by
#color(blue)("% w/w" = "mass of solute"/"mass of solution" xx 100)#
To make the calculations easier, let's pick a
#m_(NaOH) = (m_"solution" * "%w/w")/100#
#m_(NaOH) = ("100.0 g" * 20.0)/100 = "20.0 g NaOH"#
This means that the sample will contain
#m_"solution" = m_"water" + m_(NaOH)#
#m_"water" = "100.0 g" - "20.0 g" = "90.0 g water"#
Use the molar masses of sodium hydroxide and water to figure out how many moles of each you have in the sample
#20.0 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = "0.500 moles NaOH"#
#80.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.441 moles H"_2"O"#
The mole fraction of sodium hydroxide is defined as the number of moles of sodium hydroxide divided by the total number of moles present in solution.
#n_"total" = n_(NaOH) + n_"water"#
#n_"total" = "0.500 moles" + "4.441 moles" = "4.941 moles"#
This means that you have
#color(blue)(chi_"NaOH" = n_(NaOH)/n_"total")#
#chi_(NaOH) = (0.500 color(red)(cancel(color(black)("moles"))))/(4.941color(red)(cancel(color(black)("moles")))) = color(green)(0.101)#
To get the molality of the solution, you need to divide the number of moles of solute by the mass of the solvent expressed in kilograms.
#color(blue)(b = n_(NaOH)/m_"water")#
Convert the mass of water from grams to kilograms
#80.0 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 80.0 * 10^(-3)"kg"#
The molality of the solution will thus be
#b = "0.500 moles"/(80.0 * 10^(-3)"kg") = color(green)("6.25 mol kg"^(-1))#