# Converting mm Hg to mm H2O

So here's my problem:

What is the osmotic pressure, noted in mm of $\ce{H2O}$, of a solution with 125 micrograms of vitamin B12 ($\ce{C63H88CoN14O14P}$) in 2.50 mL of $\ce{H2O}$ at 25 degrees Celsius.

The "hint" that comes with the problem says: Weak osmotic pressures are often measured and given in mm of $\ce{H2O}$. Use here the density of water (1,00 g/mL) and of $\ce{Hg}$ (13,6 g/mL). In this case, the constant $R$ has a value of $\pu{62,36 mm Hg * L / mol * K}$.

Here:

$$2.50 \,\text{microgram} = 0.000125 \,\text{grams}$$

$$pi = cRT$$

$$\frac{(\pu{0.000125 g * 1 mol / 1355.3652 g})}{(\pu{2.50 mL / 1000})} * \pu{62.36 mm \ce{Hg} * L / mol * K * 298.15 K} = \pu{0.685890194 mm \ce{Hg}}$$

So, I blocked here and I saw that the solution was:

$$\pu{0.685890194 mm \ce{Hg} * 13.6 g/ml / 1.00 g/ml = 9.32 mm \ce{H2O}}$$

So, here's two questions:

• Why did I have to use R = 62.36 and not the usual R = 8.31?

• Also, what's happening in the conversion exactly? How was I supposed to figure out what they did to convert from mm $\ce{Hg}$ to mm $\ce{H2O}$?

• I wanted to make a joke but I won't do it. We need to be serious :-) – ParaH2 May 1 '17 at 21:31