Question #c6c4e

1 Answer
Aug 13, 2017

Answer:

#"molarity" = 1.19# #M#

#"molality" = 1.13# #m#

Explanation:

We're asked to find

of a solution with density #1.1# #"kg/L"# with a mole fraction of #"NaOH"# of #0.02#.

MOLALITY:

Since the mole fraction is given as #0.02#, we can set up the following relationship using the definition of mole fraction:

#"mole fraction NaOH" = (0.02color(white)(l)"mol NaOH")/(0.02color(white)(l)"mol NaOH" + 0.98color(white)(l)"mol H"_2"O") = 0.02#

Since there are proportionally #0.98# #"mol H"_2"O"#, we can use the molar mass of water to find the number of grams (and then kilograms) of water:

#0.98cancel("mol H"_2"O")((18.015cancel("g H"_2"O"))/(1cancel("mol H"_2"O")))((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = color(red)(ul(0.0177color(white)(l)"kg H"_2"O"#

We can now find the molality:

#"molality" = "mol solute"/"kg solvent" = (0.02color(white)(l)"mol NaOH")/(color(red)(0.0177color(white)(l)"kg H"_2"O")) = color(blue)(ulbar(|stackrel(" ")(" "1.13color(white)(l)m" ")|)#

#" "#

MOLARITY:

The equation for molarity is

#"molarity" = "mol solute"/"L solution"#

We can calculate the molarity from the molality if the solution's density is known (given as #1.1# #"kg/L"# #= 1.1# #"g/mL"#).

What we need to do is convert the #0.02# #"mol NaOH"# to grams (using its molar mass), and add that value to the mass of water (equal to #17.7# #"g"#) to find the total mass of solution:

#0.02cancel("mol NaOH")((40.00color(white)(l)"g NaOH")/(1cancel("mol NaOH"))) = color(red)(ul(0.800color(white)(l)"g NaOH"#

The total mass of the solution is thus

#"g solution" = color(red)(0.800color(white)(l)"g NaOH") + 17.7color(white)(l)"g H"_2"O" = color(green)(ul(18.5color(white)(l)"g solution"#

Now we use the density of the solution to calculate the number of liters:

#color(green)(18.5)cancel(color(green)("g soln"))((1cancel("mL soln"))/(1.1cancel("g soln")))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(purple)(ul(0.0168color(white)(l)"L solution"#

And the molarity is thus

#"molarity" = (0.02color(white)(l)"mol NaOH")/(color(purple)(0.0168color(white)(l)"L solution")) = color(blue)(ulbar(|stackrel(" ")(" "1.19color(white)(l)M" ")|)#