JEE Main 2025 (Online) 23rd January Evening Shift | Solutions Question 21 | Chemistry

JEE Main 2025 (Online) 23rd January Evening Shift

MCQ (Single Correct Answer)

-1

Consider a binary solution of two volatile liquid components 1 and $2 . x_1$ and $y_1$ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as :

$\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$

$\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$

$\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$

$\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$

JEE Main 2025 (Online) 22nd January Morning Shift

MCQ (Single Correct Answer)

-1

Arrange the following solutions in order of their increasing boiling points.

(i) $10^{-4} \mathrm{M} \mathrm{NaCl}$

(ii) $10^{-4} \mathrm{M}$ Urea

(iii) $10^{-3} \mathrm{M} \mathrm{NaCl}$

(iv) $10^{-2} \mathrm{M} \mathrm{NaCl}$

$($ i $)<($ ii $)<($ iii $)<($ iv $)$

(ii) $<($ i $)<($ iii $)<($ iv $)$

(iv) $<($ iii $)<($ i $)<($ ii $)$

(ii) $<$ (i) $\equiv$ (iii) $<$ (iv)

JEE Main 2024 (Online) 9th April Morning Shift

MCQ (Single Correct Answer)

-1

$$0.05 \mathrm{M} \mathrm{~CuSO}_4$$ when treated with $$0.01 \mathrm{M} \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7$$ gives green colour solution of $$\mathrm{Cu}_2 \mathrm{Cr}_2 \mathrm{O}_7$$. The two solutions are separated as shown below : [SPM : Semi Permeable Membrane]

JEE Main 2024 (Online) 9th April Morning Shift Chemistry - Solutions Question 29 English

Due to osmosis :

Molarity of $$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$$ solution is lowered.

Green colour formation observed on side Y.

Molarity of $$\mathrm{CuSO}_4$$ solution is lowered.

Green colour formation observed on side X.

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

We have three aqueous solutions of $\mathrm{NaCl}$ labelled as ' $\mathrm{A}$ ', ' $\mathrm{B}$ ' and ' $\mathrm{C}$ ' with concentration $0.1 \mathrm{M}$, $0.01 \mathrm{M}$ and $0.001 \mathrm{M}$, respectively. The value of van 't Hoff factor(i) for these solutions will be in the order :

$\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{C}}<\mathrm{i}_{\mathrm{B}}$

$\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{B}}<\mathrm{i}_{\mathrm{C}}$

$\mathrm{i}_{\mathrm{A}}>\mathrm{i}_{\mathrm{B}}>\mathrm{i}_{\mathrm{C}}$

$\mathrm{i}_{\mathrm{A}}=\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}}$

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