# Question 31c13

May 29, 2017

Here's what I got.

#### Explanation:

The information you provided here is a bit sketchy, but it does allow you to find an expression for the molality of the solution.

Molality is defined as the number of moles of solute present for every $\text{1 kg}$ of solvent.

Now, you know that you're at room temperature, so assume that the temperature is equal to ${20}^{\circ} \text{C}$ and look up the density of pure water at this temperature.

You'll find that

${\rho}_{\text{water" = "0.9982 g mL}}^{- 1}$

https://www.simetric.co.uk/si_water.htm

This means that the mass of the solvent will be equal to

100 color(red)(cancel(color(black)("mL"))) * "0.9982 g"/(1color(red)(cancel(color(black)("mL")))) = "99.82 g"

To calculate the molality of the solution, you must use the molar mass of the solute to convert its mass to moles, then determine the number of moles of solute present for every $\text{1 kg} = {10}^{3}$ $\text{g}$ of solvent.

If you take ${M}_{M}$ ${\text{g mol}}^{- 1}$ to be the molar mass of the solute, you will have

40 color(red)(cancel(color(black)("g"))) * "1 mole"/(M_Mcolor(red)(cancel(color(black)("g")))) = (40/M_M) $\text{moles}$

The number of moles of solute present for every ${10}^{3}$ $\text{g}$ of solvent will be

10^3 color(red)(cancel(color(black)("g water"))) * ((40/M_M)color(white)(.)"moles")/(99.82color(red)(cancel(color(black)("g water")))) = (400.72/M_M)# $\text{moles}$

Therefore, the solution has a molality of

$\text{molality} = \left(\frac{400.72}{M} _ M\right)$ ${\text{mol kg}}^{- 1}$

Keep in mind that you must round the actual value to one significant figure, the number of sig figs you have for your values.