Question

If $20\;g$ of a solute was dissolved in $500\;mL$ of water and osmotic pressure of the solution was found to be $600mm\;of\;Hg$ at ${{15}^\circ}C$ , then molecular weight of the solute is:
A) $1000\;$
B) $1200\;$
C) $1400\;$
D) $1800\;$

Answer 1

Hint: We can find the molecular weight of the solute by using the Law of osmotic pressure which explains the osmotic pressure of a solute as the product of molar concentration, universal gas constant and the temperature of the solute. The value of the gas constant will be taken as per the unit of pressure and volume.

Complete answer:
Let us note down the given data and data we need to find;
Mass of the solute (in $g$ ) $m=20g$
Volume of the solute (in $L$ ) $V=500mL=0.5L$
Osmotic pressure of the solute (in $mm\;of\;Hg$ ) $p=600mm\;of\;Hg$
Temperature of solute (in $K$ ) $T=15{}^\circ C=288K$
Value of Universal Gas constant (in $LatmK^{-1}mol^{-1}$ ) $R=0.0821Latm{{K}^{-1}}mo{{l}^{-1}}$
Molar concentration of solute $C=?$
Moles of the solute (in $\;mol$ ) $n=?$
Molecular weight of solute (in $g$ ) $M=?$
Here, we are given the pressure in terms of $mm\;of\;Hg$ . We know that the standard value
$760mm\;of\;Hg=1atm$
Thus, the given pressure can be converted to atmospheric pressure as
$ p=\dfrac{600}{760}=0.7895atm$
Now, the osmotic pressure can be calculated by the formula
$p=CRT$
Substituting the given values,
$ 0.7895=C\times 0.0821 \times 288$
$\therefore C=0.0334mol{{L}^{-1}}$
Now, the molar concentration can be defined as the ratio of the moles of the solute to the volume of the solute
$C=\dfrac{n}{V}$
Substituting the provided values;
$ 0.0334=\dfrac{n}{0.5}$
$n=0.0167mol$
Now, the moles of the solute is the given mass of the solute per molar mass of the solute.
$ n=\dfrac{m}{M}$
Substituting the obtained values,
$ 0.0167=\dfrac{20}{M}$
$ M\simeq 1200g$

Hence, the correct answer is $(B)$


Note: We should always select the value of universal gas constant by considering the unit in which the pressure and volume are given, as it varies with the units. We should remember the value of gas constant in two standard units i.e. $R=0.0821Latm{{K}^{-1}}mo{{l}^{-1}}$ and $R=8.31J{{K}^{-1}}mo{{l}^{-1}}$ . We also have the value of gas constant in the unit of height of mercury, but as it is not given we convert it to atmospheric pressure, as the value of gas constant in terms of atm. pressure is to be remembered.