Question

(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass $=180g{mol}^{-1}$) in water is labeled as $10$% (by mass). What would be the molality and molarity of the solution? (Density of solution $=1.2g{mL}^{-1}$)

Answer 1

(a) Positive deviation

Let Take a binary solution with have components A and B. If the force of attraction between molecular of and A and B in the solution are weaker than that of between A — A and B — B, then the tendency of escaping of molecules A—B from the solution becomes more than that of pure liquids. The total pressure of the solution will be greater than the corresponding vapour pressure of ideal solution of the same component A and B. This type of solution shows positive deviation from Raoult's law

(b) 10% w/w solution of glucose in water that 10g of glucose in 100 g of the solution i.e, 10g of glucose is present in (100-10) -90g of water.

Molar mass of glucose $\left(C_6{H}_{12}{O}_6\right)=6\times 12+12\times 1+6\times 16=180gmo{l}^{-1}$

Then, number of moles of glucose = 10/180 mol = 0.056 mol

$\therefore$ Molality of solution = 0.056 mol / 0.09 kg = 0.62m

Number of moles of water $=90g/18gmo{l}^{-1}=5mol$

Mole fraction of glucose $\left(x_g\right)=0.056/(0.056+5)=0.011$

And, mole fraction of warer $x_w=1-x_g$

$=1-0.011=0.989$

If the density of the solution is 1.2gm$L^{-1}$, then the volume of the 100 g solution can be given as:

$=100g/1.2gm{L}^{-1}$

$=83.33mL$

$=83.33\times {10}^{-3}L$

$\therefore$ Molarity of the solution $=0.056mol/83.33\times {10}^{-3}L$

$=0.67M$