Solutions and Colligative Properties - Result Question 66
Passage
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life.
One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is $0.9$ .
Given, freezing point depression constant of water
$ \left(K_f^{\text {water }}\right)=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} $
Freezing point depression constant of ethanol
$ \left(K_f^{\text {ethanol }}\right)=2.0 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} $
Boiling point elevation constant of water
$ \left(K_b^{\text {water }}\right)=0.52 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} $
Boiling point elevation constant of ethanol
$ \left(K_b^{\text {ethanol }}\right)=1.2 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} $
Standard freezing point of water $=273 \mathrm{K}$
Standard freezing point of ethanol $=155.7 \mathrm{K}$
Standard boiling point of water $=373 \mathrm{K}$
Standard boiling point of ethanol $=351.5 \mathrm{K}$
Vapour pressure of pure water $=32.8 \mathrm{mm} $ $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{mm} $ $\mathrm{Hg}$
Molecular weight of water $ =18 $ $\mathrm{g} $ $\mathrm{mol}^{-1}$
Molecular weight of ethanol $=46$ $ \mathrm{g}$ $ \mathrm{mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
$(2008,3 \times 4 M=12 M)$
37. Water is added to the solution $M$ such that the mole fraction of water in the solution becomes $0.9$ . The boiling point of this solution is
(a) $380.4$ $ K$
(b) $376.2$ $ K$
(c) $375.5$ $ K$
(d) $354.7 $ $K$
Show Answer
Answer:
Correct Answer: 37. (b)
Solution:
Now ethanol is solute.
Molality of solute $=\frac{0.1}{0.9 \times 18} \times 1000=6.17$
$\Rightarrow \Delta T _b =6.17 \times 0.52=3.20 $
$\Rightarrow T _b =373+3.2 $
$=376.2 K$
BETA
