# Question 069c0

Sep 18, 2017

#### Answer:

Here's what I got.

#### Explanation:

As you know, the osmotic pressure of a solution can be calculated using the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Pi = i \cdot c \cdot R T}}}$

Here

• $\Pi$ is the osmotic pressure of the solution
• $i$ is the van't Hoff factor
• $c$ is the molarity of the solution
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
• $T$ is the absolute temperature of the solution

Now, the problem tells you that the solute is sucrose, a nonelectrolyte, i.e. a compound that does not dissociate in aqueous solution.

This means that the van't Hoff factor, which tells you the ratio of moles of particles of solute produced in solute per mole of particles of solute dissolved in solution, is equal to $1$.

So, you know that your solution has a molality equal to $\text{2 molal}$, which means that you get $2$ moles of solute for very $\text{1 kg}$ of solvent.

To make the calculations easier, let's pick a sample of this solution that contains exactly $\text{1 kg} = {10}^{3}$ $\text{g}$ of water. Use the molar mass of sucrose to find the mass of sucrose present in the sample

2 color(red)(cancel(color(black)("moles sucrose"))) * "342.3 g"/(1color(red)(cancel(color(black)("mole sucrose")))) = "684.6 g"

You can thus say that the total mass of the solution will be

$\text{1000 g + 684.6 g = 1684.6 g}$

Use the density of the solution to calculate its volume

1684.6 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.4color(red)(cancel(color(black)("g")))) = "1203.3 mL"

Now, in order to find the molarity of the solution, you must find the number of moles of solute present in exactly ${10}^{3} \textcolor{w h i t e}{.} \text{mL" = "1 L}$ of this solution.

To do that, use the fact that $\text{1203.3 mL}$ contain 2 moles of sucrose.

10^3 color(red)(cancel(color(black)("mL solution"))) * "2 moles sucrose"/(1203.3color(red)(cancel(color(black)("mL solution")))) = "1.662 moles sucrose"

Therefore, you can say that you have

$c = {\text{1662 mol L}}^{- 1}$

You can now plug in your values into the equation and solve for the osmotic pressure of the solution--do not forget that you must use the fact that

$T \left[\text{K"] = t[""^@"C}\right] + 273.15$

to convert the temperature from degrees Celsius to Kelvin!

$\Pi = 1 \cdot 1.662 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"))) color(red)(cancel(color(black)("L"^(-1)))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (25 + 273.15)color(red)(cancel(color(black)("K}}}}$

$\Pi = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{41 atm}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the molality of the solution.