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 $\ce{H2O}$?
You CAN use R=8.31 but it has the (standard) unit of (m3 Pa K−1 mol−1), simply convert your volume from liter (aka dm3) to meter cube (m3) and the pressure will be in Pascal (Pa)

If you use R=62.36, which has a unit of (L mmHg K−1   mol−1), as long as your   volume is in liter, the pressure will be in mmHg
But the question ask for mmH2O.

Pressure is simply force on an area (F/A), weight on an area (mg/A), density * volume on a area (rho*V /A) = density * height (rho*h)

Thus P = density (Hg) * h (Hg) = density (water) * h (water)

h (water) =  (13.6 / 1.00) * h(Hg)