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# DETERMINATION OF MOLECULAR MASSES

In the case of dilute solutions, it has been stated that the equation PV = nST holds good. Instead of one gram mole of the solute present in V litres of solution, let wAgram of solute (mol. mass mA be present in V’ litres of solution; then)

n = wA/mA and V = V'

Thus, the equation PV = nST becomes

PV' = wA/mA.ST

or     mA = wA×S×T/PV'

Knowing the values of P experimentally, the value of mA, i.e., molecular mass of the solute can be determined.

Consider two solutions I and II having n1 and n2 moles of the solute in V1 and V2litres of solution respectively. Let P1 and P2 be their osmotic pressures at the same temperature (T).

From the equation       PV = nST

For solution I,             P1V1 = n1ST

or                             p1 = n1/V1 ST

For solution II,            P2V2 = n2ST

or                             p2 = n2/V2 ST

If both solutions are isotonic, i.e., P1 = P2, obviously,

n1/V1 ST = n2/V2 ST

or       n1/V1 = n2/V2

or       wa/m1/V1 = w2/m2/V2

or       wq/m1×V1 = w2/m2×V2

This is the condition for isotonic solutions.

If molecular mass of one solute is known, the molecular mass of the other can be determined without using osmotic pressure values.

Osmotic pressure of mixture of two solutions:

Case I: Let two solutions of same substance having different osmotic pressure π and π are mixed. Osmotic pressure of the resultant solution can be calculated as,

π1V1 = π2V2 = πR(V1 + V2)

where, V1 and V2 are the volumes of two solutions and πR is the resultant osmotic pressure.

Case II: Let n1 and n2 are the number of moles of two different solutes present in V1 and V2 volumes respectively.

Osmotic pressure of the mixture can be calculated as,

π = π1 + π2 = n1i2RT/(V1+V2) + n2i2RT/(V1+V2)

π = (n1i1 + n2i2)/(V1+V2) RT

Here, i1 and i2 are van’t Hoff factor for the two solutes.

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