# Question 71376

Nov 8, 2015

(C) $\text{0.70 molal}$

#### Explanation:

I'll show you how to solve part (C), sicne that is the most interesting of the three solutions, and you can solve the first two parts as practice.

Molality is defined as moles of solute per kilograms of solvent.

$\textcolor{b l u e}{\text{molality" = "moles of solute"/"kilograms of solvent}}$

This means that in any problem that asks you to find the molality of a given solution, you need to determine two things

• the number of moles of solute
• the mass of the solute - expressed in kilograms!

In your case, you are given the number of moles of solute and the mass of the solvent. Notice, however, that this value is given in grams, which automatically tells you that you need to convert it to kilograms before using it in the equation.

Use the known conversion factor that exists between grams and kilograms

${10}^{3} \text{g " = " 1 kg}$

to get

20.1color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 20.1 * 10^(-3)"kg"#

Now plug in your values and solve for the molaity of the solution

$b = {n}_{'} s o l u t e \text{/m_"solvent}$

$b = \text{0.014 moles"/(20.1 * 10^(-3)"kg") = "0.6965 molal}$

You need to round this off to two sig figs, the number of sig figs you have for the number of moles of solute

$b = \textcolor{g r e e n}{\text{0.70 molal}}$