1
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?

A
0.2
B
0.4
C
0.8
D
0.6
2
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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 :

A
$\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$
B
$\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$
C
$\frac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \frac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}$
D
$\frac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \frac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}$
3
JEE Main 2025 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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

A
$($ i $)<($ ii $)<($ iii $)<($ iv $)$
B
(ii) $<($ i $)<($ iii $)<($ iv $)$
C
(iv) $<($ iii $)<($ i $)<($ ii $)$
D
(ii) $<$ (i) $\equiv$ (iii) $<$ (iv)
4
JEE Main 2024 (Online) 9th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$0.05 \mathrm{M} \mathrm{~CuSO}_4$$ 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 :

A
Molarity of $$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$$ solution is lowered.
B
Green colour formation observed on side Y.
C
Molarity of $$\mathrm{CuSO}_4$$ solution is lowered.
D
Green colour formation observed on side X.
JEE Main Subjects
EXAM MAP