# Question #069c0

##### 1 Answer

Here's what I got.

#### Explanation:

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

#color(blue)(ul(color(black)(Pi = i * c * R T)))#

Here

#Pi# is theosmotic pressureof the solution#i# is thevan't Hoff factor#c# is themolarityof the solution#R# is theuniversal gas constant, equal to#0.0821("atm" * "L")/("mol" * "K")# #T# is theabsolute temperatureof 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

So, you know that your solution has a **molality** equal to **moles** of solute for very **solvent**.

To make the calculations easier, let's pick a sample of this solution that contains exactly **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

#"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

To do that, use the fact that **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 = "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["K"] = t[""^@"C"] + 273.15#

to convert the temperature from *degrees Celsius* to *Kelvin*!

#Pi = 1 * 1.662 color(red)(cancel(color(black)("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 = color(darkgreen)(ul(color(black)("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.