Chemistry - Liquid Solution Question with Solution | TestHub

Question Explanation: Match the column. Concept: This question is based on the van't Hoff factor. Solution:

$$M_{observed}=\frac{M_{theoretical}}{i}$$

For ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}$, since it dissociates, the number of particles increases, so $i>1$. Therefore, $M_{observed}<M_{theoretical}$.

P → (1)

(Q) $0.1\mathrm{M}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ (glucose): Glucose is a non-electrolyte. It dissolves in water but does not dissociate or associate. The number of particles remains the same, so $i=1$. Therefore, $M_{observed}=M_{theoretical}$.

Q → (2)

(R) $0.05\mathrm{M}$ Barium chloride in water: Barium chloride (${\mathrm{BaCl}}_2$) is a strong electrolyte and dissociates completely in water, producing three ions.

$${\mathrm{BaCl}}_2\to {\mathrm{Ba}}^{2+}+2{\mathrm{Cl}}^{-}$$

The number of particles becomes three times the initial amount, so $i=3$. Therefore,

$$M_{observed}=\frac{M_{theoretical}}{3}$$R → (4)

(S) $0.1\mathrm{M}{\mathrm{CH}}_3\mathrm{COOH}$ in benzene:

In a non-polar solvent like benzene, acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$) molecules associate via hydrogen bonding to form dimers.

$2{\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons ({\mathrm{CH}}_3\mathrm{COOH}{)}_2$

Since two molecules combine to form one, the number of particles decreases, so

$i<1$. Therefore, $M_{observed}>M_{theoretical}$. S → (3)

Final Answer: option (B)