1
JEE Main 2024 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

Considering acetic acid dissociates in water, its dissociation constant is $$6.25 \times 10^{-5}$$. If $$5 \mathrm{~mL}$$ of acetic acid is dissolved in 1 litre water, the solution will freeze at $$-x \times 10^{-2}{ }^{\circ} \mathrm{C}$$, provided pure water freezes at $$0{ }^{\circ} \mathrm{C}$$.

$$x=$$ _________. (Nearest integer)

$$\begin{aligned} \text{Given :} \quad & \left(\mathrm{K}_{\mathrm{f}}\right)_{\text {water }}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}-1 \\ & \text { density of acetic acid is } 1.2 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of water }=18 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of acetic acid= } 60 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { density of water }=1 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned}$$

Acetic acid dissociates as $$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{\oplus}$$

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2
JEE Main 2024 (Online) 5th April Morning Shift
Numerical
+4
-1
Change Language

An artificial cell is made by encapsulating $$0.2 \mathrm{~M}$$ glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a $$0.05 \mathrm{~M}$$ solution of $$\mathrm{NaCl}$$ at $$300 \mathrm{~K}$$ is ________ $$\times 10^{-1}$$ bar. (nearest integer).

[Given : $$\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$$ ]

Assume complete dissociation of $$\mathrm{NaCl}$$

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3
JEE Main 2024 (Online) 4th April Evening Shift
Numerical
+4
-1
Change Language

$$2.7 \mathrm{~kg}$$ of each of water and acetic acid are mixed. The freezing point of the solution will be $$-x^{\circ} \mathrm{C}$$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $$x=$$ ________ (nearest integer)

[Given: Molar mass of water $$=18 \mathrm{~g} \mathrm{~mol}^{-1}$$, acetic acid $$=60 \mathrm{~g} \mathrm{~mol}^{-1}$$

$${ }^{\mathrm{K}_{\mathrm{f}}} \mathrm{H}_2 \mathrm{O}: 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$

$$\mathrm{K}_{\mathrm{f}}$$ acetic acid: $$3.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$

freezing point: $$\mathrm{H}_2 \mathrm{O}=273 \mathrm{~K}$$, acetic acid $$=290 \mathrm{~K}$$]

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4
JEE Main 2024 (Online) 4th April Morning Shift
Numerical
+4
-1
Change Language

$$2.5 \mathrm{~g}$$ of a non-volatile, non-electrolyte is dissolved in $$100 \mathrm{~g}$$ of water at $$25^{\circ} \mathrm{C}$$. The solution showed a boiling point elevation by $$2^{\circ} \mathrm{C}$$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________ $$\mathrm{mm}$$ of $$\mathrm{Hg}$$ (nearest integer)

[Given : Molal boiling point elevation constant of water $\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}$, $$1 \mathrm{~atm}$$ pressure $$=760 \mathrm{~mm}$$ of $$\mathrm{Hg}$$, molar mass of water $$=18 \mathrm{~g} \mathrm{~mol}^{-1}]$$

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