Solutions and Colligative Properties - Result Question 63
####34. Addition of $0.643 g$ of a compound to $50 mL$ of benzene (density : $0.879 g / mL$ ) lowers the freezing point from $5.51^{\circ} C$ to $5.03^{\circ} C$. If $K _f$ for benzene is 5.12 , calculate the molecular weight of the compound.
$(1992,2 M)$
Passage Based Questions
Passage 1
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
$$ \left(K _f^{\text {water }}\right)=1.86 K kg mol^{-1} $$
Freezing point depression constant of ethanol
$$ \left(K _f^{\text {ethanol }}\right)=2.0 K kg mol^{-1} $$
Boiling point elevation constant of water
$$ \left(K _b^{\text {water }}\right)=0.52 K kg mol^{-1} $$
Boiling point elevation constant of ethanol
$$ \left(K _b^{\text {ethanol }}\right)=1.2 K kg mol^{-1} $$
Standard freezing point of water $=273 K$
Standard freezing point of ethanol $=155.7 K$
Standard boiling point of water $=373 K$ Standard boiling point of ethanol $=351.5 K$
Vapour pressure of pure water $=32.8 mm Hg$
Vapour pressure of pure ethanol $=40 mm Hg$
Molecular weight of water $=18 g mol^{-1}$
Molecular weight of ethanol $=46 g mol^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
$(2008,3 \times 4 M=12 M)$
Show Answer
Solution:
- $-\Delta T _f=5.51-5.03=0.48$
$\Rightarrow \quad-\Delta T _f=0.48=K _f \cdot m$
$\Rightarrow \quad 0.48=5.12 \times \frac{0.643}{M} \times \frac{1000}{50 \times 0.879}$
$\Rightarrow \quad M=156 g / mol$
