Question #9e925

The osmotic pressure of an aqueous solution can be calculated by using the equation

`color(blue)(ul(color(black)(Pi = i * c * RT)))`

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, usually given as `0.0821("atm" * "L")/("mol" * "K")`
• `T` is the absolute temperature of the solution

Now, you didn't provide a value for the temperature of the solution, so I'll just assume that you're working at room temperature

`T = 20^@"C" + 273.15 = "293.15 K"`

You know that the van't Hoff factor is said to be equal to `1`, so you can rewrite the above equation as

`Pi = c * RT`

Rearrange to solve for `c`

`c = Pi/(RT)`

Plug in your values to find

`c = (0.037 color(red)(cancel(color(black)("atm"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 293.15color(red)(cancel(color(black)("K")))) = "0.001537 mol L"^(-1)`

Use the molarity and the volume of the solution, which you can assume to be equal to the volume of water, to determine the number of moles of solute present in the solution

`0.200 color(red)(cancel(color(black)("L solution"))) * "0.001537 moles solute"/(1color(red)(cancel(color(black)("L solution")))) = "0.0003074 moles solute"`

To find the molar mass of the protein, use the fact that `"5.00 g"` of this protein contain `0.0003074` moles

`1 color(red)(cancel(color(black)("mole protein"))) * "5.00 g"/(0.0003074color(red)(cancel(color(black)("moles protein")))) = "16,265 g"`

Since the molar mass of the protein tells you the mass of `1` mole of this protein, you can say that you will have

`color(darkgreen)(ul(color(black)(M_"M protein" = "16,300 g mol"^(-1))))`

I'll leave the answer rounded to three sig figs, but keep in mind that you only have two sig figs for the osmotic pressure.