*A* and *B* solutions has been made using a solid organic compound called *X*. In *A* there are 5g of *X* and 100g of *water*. In *B* there are 2.3g of *X* and 100g of *Benzene*. In related boiling points vapour pressure of pure water and pure Benzene is 101300 Pa a piece. Vapour pressure of *A* and *B* In the same temperature is 100570 Pa a piece. Find the relative molecular weight of X in each solution and explain why it is different each other.

(Note : You can use P^o - P/P^o equation)

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When a non-volatile solid is dissolved in a solvent, the vapour pressure of the resulting solution becomes less than that of the pure solid. The lowering of vapour pressure due to dissolution can be obtained from Rauolt’s law which states that, the relative lowering of vapour pressure is equal to the mole-fraction of the solute.

Or, `(deltap)/P^0=x_2`

In solution A, mole fraction of X (let, its molar mass be M)

=`(5/M)/(5/M+100/18)`

Plugging in the values in Raoult’s law equation,

`(101300-100570)/101300 =(5/M)/(5/M+100/18)`

Upon solving we get, M =124 g/mole

Again, considering solution B, we can write,

`(101300-100570)/101300 =(2.3/W)/(2.3/W+100/78)`

Upon solving we get, W =106.4 g/mole

So, the two solutions of the same organic compound returns two values of molar mass. This is not uncommon while studying colligative properties of dilute solutions, since the substances often behave in unusual ways in solutions. Dissociation or association of molecules occur in solution, under certain conditions. The results obtained above reveal that either the compound is in an associated state in water, or it has undergone dissociation in benzene, or both.

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Thak you very much

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