Question

Question #c6c4e

Answer 1

`"molarity" = 1.19` `M`

`"molality" = 1.13` `m`

Explanation:

We're asked to find

• the molarity

• the molality

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" ")|)`