The height of a column of liquid that can be supported by a given pressure is inversely proportional to its density. An aqueous solution of $0.010 g$ of a protein in $10 mL$ of water at $20^oC$ shows a rise of $5.22 cm$ in a mercury tube. Assume the density of the solution to be $0.998 g/cm^3$ and the density of the mercury to be $13.6 g/cm^3$. (a) What is the molar mass of the protein?

The answer key says the answer is $4.8*10^3 g/mol$, but I got $3.5*10^2 g/mol$. Below is my reasoning:

Osmotic pressure = pressure to raise mercury, so $MRT = rho * g * h$

$\frac{(0.010/M)}{(10.010/0.998/1000)} * 8.3145 * 293 * 10^3$ (pascals) = $13.6/1000*(100)^3 * 9.8 * 0.0522$ (pascals). Solving this equation for $M$ gets $3.49*10^2$, instead of 3.8*10^3.

Can someone please point out where I am wrong?

Please do not point this out as a duplicate, as I have already referenced the question here. I used their approach but still cannot get the correct answer.

  • $\begingroup$ Osmotic pressure is $cRT$, not $MRT$. You should put all dimension in your equation, and make sure they check out! $\endgroup$ – Karl Jul 30 '18 at 5:36
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    $\begingroup$ If you are going to use R in the SI units so, do the same for the rest, volume and mass. As Karl said, putting all dimensions in the calculation helps you to identify the mistakes, because in the end it must give you a molar mass unit like $kg mol^{-1}$. Please format your equation with the frac{}{} command and dont froget the \ on rho, if is convinient is better use two $ than just one for the solution. Ah! and the link doesn't work, plese check it. $\endgroup$ – liuzp Jul 30 '18 at 6:05
  • $\begingroup$ Just to make this clear, a 1 M solution has a concentation $c = 1 \rm{mol/l}$. "M" for "molar" is not a mathematical symbol. The capital $M$ designates molar mass, which is the number you want. $\endgroup$ – Karl Jul 30 '18 at 18:52
  • $\begingroup$ @liuzp While we're nitpicking, units should be given in roman, not italics, by using \rm{} ;-) $\endgroup$ – Karl Jul 30 '18 at 18:56
  • $\begingroup$ @Karl. Since when does M designate molar mass and not moles/liter? I've seen it used to represent molar mass (MW), but most often it seems to be used to express concentration in moles/liter (molarity). $\endgroup$ – Dr. J. Jul 30 '18 at 21:15

Osmotic pressure of a solute in water is given by the formula

$$\Pi = i\rm{MRT}\tag{1}$$

$\Pi$ is the osmotic pressure
$i$ is the Van't Hoof factor
$\rm{M}$ is the molarity of the solute (mol/L)
$\rm{R}$ is the universal gas constant 62.36 ($\rm{L}\space \rm{Torr}\space$ $\rm{K}^{−1}$ $\rm{mol}^{−1}$)
$\rm{T}$ is the absolute temperature ($^\circ\rm{K}$)

We'll assume $i =1$.

Let's use $\rm{M_W}$ for the molecular weight of the protein. So the moles of the protein is equal to $\rm{g/M_W}$.

$$\dfrac{0.010\space\rm{g}/\rm{M_W}}{0.010\space\rm{L}}\times 62.36\space\rm{L}\space \rm{Torr}\space\rm{K}^{−1}\space\rm{mol}^{−1}\times293\space\rm{K} = 52.2 \space \rm{Torr}\tag{2}$$

$$\rm{M_W} = \dfrac{0.010\times62.36\times293}{0.010\times52.2}\space\rm{g/mol}= 350\space\rm{g/mol}\tag{3} $$

Your mistake was believing the book answer.

If you need to check this sort of answer, then choose R with convenient units. It makes solving the problem much easier.

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    $\begingroup$ You (like the OP) are using M for molarity and molar mass. The former is a bit old fashioned, and it's all confusing. $\endgroup$ – Karl Jul 30 '18 at 21:11
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    $\begingroup$ @MaxW Thanks for your response. Don't you mean 0.010 L instead of 0.0010 L though? $\endgroup$ – coder Jul 31 '18 at 1:49
  • $\begingroup$ @coder - Yes, you are correct. 10 ml = 0.010 Liters. blush... Thanks for pointing out the error. $\endgroup$ – MaxW Jul 31 '18 at 2:33

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