# Question 24c18

Jul 4, 2017

a) Molarity = 4.3 mol/L; b) molality = 5.4 mol/kg.

#### Explanation:

a) Molarity

The formula for the molarity $c$ of a solution is

color(blue)(bar(ul(|color(white)(a/a)c = "Moles"/"Litres" = n/Vcolor(white)(a/a)|)))" "

Assume that you have 100 g of solution.

Then you have 20 g of ethanol.

$\text{Moles of ethanol" = 20 color(red)(cancel(color(black)("g EtOH"))) × "1 mol EtOH"/(46.07 color(red)(cancel(color(black)("g EtOH")))) = "0.434 mol EtOH}$

V = 100 color(red)(cancel(color(black)("g solution")))× "1 mL solution"/(0.996 color(red)(cancel(color(black)("g solution")))) = "100.4 mL solution" = "0.1004 L solution"

$c = \text{0.434 mol"/"0.1004 L" = "4.3 mol/L}$

b) Molality

The formula for the molality $b$ of a solution is

color(blue)(bar(ul(|color(white)(a/a)b = "moles of solute"/"kilograms of solvent" color(white)(a/a)|)))" "#

Assume that you have 100 g of solution.

Then you have 20 g of ethanol and 80 g of water.

$\text{Moles of ethanol = 0.434 mol EtOH}$

$\text{Kilograms of H"_2"O" = 80 color(red)(cancel(color(black)("g H"_2"O"))) × ("1 kg H"_2"O")/(1000 color(red)(cancel(color(black)("g H"_2"O")))) = "0.0800 kg H"_2"O}$

$b = \text{0.434 mol"/"0.0800 kg" = "5.4 mol/kg}$