Question #aaf98

1 Answer
Feb 2, 2016

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 #"20.0% w/w"# sodium hydroxide solution and use its percent concentration by mass to figure out

  • 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 #100#

#color(blue)("% w/w" = "mass of solute"/"mass of solution" xx 100)#

To make the calculations easier, let's pick a #"100.0-g"# sample of this sodium hydroxide solution. This sample will contain

#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))#