Question

1 gm of polymer having molar mass 1,60,000 gm dissolves in 800 ml water, so calculate the osmotic pressure on pascal at $27^\circ {\rm{C}}$.
(a) 0.78
(b) 0.90
(c) 19.4
(d) 12.3

Answer 1

Hint: We know that osmosis is the process of flow of solvent through a semipermeable membrane. The flow of solvent continues till the equilibrium is attained. Here, we have to use the formula of osmotic pressure.

Complete step by step answer:
Let's understand osmotic pressure in detail. On applying some pressure the flow of solvent from its side to the solution side through a membrane (semipermeable) can be stopped. The pressure that inhibits the flow of solvent is termed as osmotic pressure. The formula of osmotic pressure is,

$\pi = CRT$…… (1)

Here, $\pi $ is osmotic pressure, C is concentration, R is gas constant and T is temperature.

Now, come to the question. Mass of the polymer (solute) is 1 gm, the molar mass of the polymer (solute) is 1,60,000 gm and the volume of solution is 800 ml.

Now, we have to calculate the concentration of the solution. The formula is,

${\rm{Concentration}} = \dfrac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{solution}}\left( {{\rm{in}}\,{\rm{L}}} \right)}}$

${\rm{Concentration}} = \dfrac{{\dfrac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Molar mass}}\,{\rm{of}}\,{\rm{solute}}}}}}{{{\rm{Volume}}\,{\rm{in}}\,{\rm{litre}}\,}}$

Now, put all the values in the above equation to calculate concentration.

${\rm{Concentration}} = \dfrac{1}{{160000}} \times \dfrac{{1000}}{{800}}$


The value of gas constant R is ${\rm{8}}.{\rm{314}} \times {10^3}\,{\rm{L}}\,{\rm{Pa}}\,\,{{\rm{K}}^{ - 1}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$
 and the temperature is given as $27^\circ {\rm{C}} = {\rm{27 + 273}} = {\rm{300}}\,{\rm{K}}$.
Now, we have to put the value of concentration, gas constant and temperature in equation (1).

$\pi = CRT$

$ \Rightarrow \pi = \dfrac{1}{{160000}} \times \dfrac{{1000}}{{800}} \times 8.314 \times {10^3} \times 300\,$

$ \Rightarrow \pi = \dfrac{{2494.2}}{{128}} = 19.4\,{\rm{Pa}}$

Therefore, osmotic pressure of the solution is 19.4 Pa.

So, the correct answer is Option c.

Note: Osmotic pressure measurement provides an alternate method to determine molar mass of solutes. In calculation of molar mass of polymers, proteins and other macromolecules this method is used. The advantage of using this method is that the pressure is measured at around room temperature.