# Question c6c4e

Aug 13, 2017

$\text{molarity} = 1.19$ $M$

$\text{molality} = 1.13$ $m$

#### Explanation:

of a solution with density $1.1$ $\text{kg/L}$ with a mole fraction of $\text{NaOH}$ 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 $0.98$ $\text{mol H"_2"O}$, 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" ")|)

$\text{ }$

MOLARITY:

The equation for molarity is

$\text{molarity" = "mol solute"/"L solution}$

We can calculate the molarity from the molality if the solution's density is known (given as $1.1$ $\text{kg/L}$ $= 1.1$ $\text{g/mL}$).

What we need to do is convert the $0.02$ $\text{mol NaOH}$ to grams (using its molar mass), and add that value to the mass of water (equal to $17.7$ $\text{g}$) 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

$\text{g solution" = color(red)(0.800color(white)(l)"g NaOH") + 17.7color(white)(l)"g H"_2"O" = color(green)(ul(18.5color(white)(l)"g solution}$

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" ")|)#