91Ó°ÊÓ

We are given a table,

For dilute solutions, the osmotic pressure can be approximated as,

$\mathrm{Ï€}=\frac{\mathrm{nRT}}{\mathrm{V}}$

Therefore,

$M=\frac{cRT}{\mathrm{Ï€}}$

Here the number density of the solute $\left(\frac{n}{V}\right)$in moles/liter is replaced by $\left(\frac{c}{M}\right)$, $c$is the concentration of the solute in (grams/liter) and $M$is its molecular weight (in grams/mole). But in this experiment the osmotic pressure is balanced by the difference in the fluid level $∆h$so for each solute concentration, one can calculate the osmotic pressure by the following formula,

$\mathrm{Ï€}=\mathrm{Òϲ\mu }∆\mathrm{h}$

Substitute $\mathrm{ÒÏ}g∆h$for $\mathrm{Ï€}$in the equation $M=\frac{cRT}{\mathrm{Ï€}}$

$$\begin{gathered} \mathrm{Ï€}=\frac{\mathrm{nRT}}{\mathrm{V}} \\ =\frac{\mathrm{cRT}}{\mathrm{M}} \end{gathered}$$

Use the density of water $\left(\mathrm{ÒÏ}=1g/c{m}^3\right)$. The table given below shows the value of the osmotic pressure for each solute condition and the corresponding estimated value of the molecular weight from the following relation,

The table is as follows,

The graph of concentration versus molecular mass is as follows,

The molecular weight estimated from each measurement is plotted versus the concentration. As the equation that relates osmotic pressure to the concentration of solutes is most accurate in the limit of very dilute solution, one should extrapolate the results to zero concentration. If apparently one bad point at $c=16.6g/L$is ignored, the molecular weight is $M=66.3kg/mol$. Therefore, the approximate molecular weight is$66.3kg/mol$