Question

5 gm of a substance with molecular weights 200 is dissolved in 50 gm of a solvent with molecular weight of 60 and vapour pressure 40 cm Hg at 27°c. Calculate the vapour pressure of the solution at this temperature. [ISC 1998]

Answer 1

Answer:

vapour pressure at 27c is 38.833 cm Hg

Explanation:

• solution is

Answer 2

Answer:

The vapour pressure of the solution at this temperature is 38.33 cm Hg.

Explanation:

Given,

weight of the substance (w1) = 5 gm

G.M.W of the substance (G.M.W1) = 200 gm

So, no. of moles of the substance (n1) = w1/G.M.W1 = 5/200 = 0.025

Similarly,

weight of the solvent (w2) = 50 gm

G.M.W of the solvent (G.M.W2) = 60 gm

So, the no. of moles of the solvent (G.M.W2) = w2/G.M.W2 = 50/60 = 0.83

And also the pure vapour pressure of the solvent ($P_2^o$) = 40 cm Hg

To find,

The vapour pressure of the solution ($P_2$)

We know that from Raoult' s law

$P_2 = \chi_2\times P_2^o$

$P_2 = \frac{n_2}{n_1+n_2} P_2^o$ (As  $\chi_2 = \frac{n_2}{n_1+n_2}$ )

$P_2 = \frac{0.83}{0.83+0.025} \times 40 cm Hg$

$P_2 = 38.33 cmHg$

Therefore, the vapour pressure of the solution at this temperature is 38.33 cm Hg.

#SPJ3