1
JEE Main 2025 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
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-1
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$X Y$ is the membrane/partition between two chambers 1 and 2 containing sugar solutions of concentration $c_1$ and $c_2\left(c_1>c_2\right) \mathrm{mol} \mathrm{L}^{-1}$. For the reverse osmosis to take place identify the correct condition.

(Here $p_1$ and $p_2$ are pressures applied on chamber 1 and 2 ).

JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Solutions Question 1 English

A. Membrane/Partition : Cellophane, $\mathrm{p}_1>\pi$

B. Membrane/Partition : Porous, $\mathrm{p}_2>\pi$

C. Membrane/Partition : Parchment paper, $p_1>\pi$

D. Membrane/Partition : Cellophane, $\mathrm{p}_2>\pi$

Choose the correct answer from the option given below:

A
A and C Only
B
A and D Only
C
B and D Only
D
C Only
2
JEE Main 2025 (Online) 4th April Morning Shift
Numerical
+4
-1
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The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_a=4 \times 10^{-10}\right)$ is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as $x \times 10^{-4} \mathrm{M}$. The value of $x$ is _________ (nearest integer)

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3
JEE Main 2025 (Online) 4th April Morning Shift
Numerical
+4
-1
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In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _______ % (nearest integer)

(Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{~Hg}$ )

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4
JEE Main 2025 (Online) 4th April Morning Shift
Numerical
+4
-1
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Fortification of food with iron is done using $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_4$. $7 \mathrm{H}_2 \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is ______ (Nearest integer)

[Given: Molar mass of $\mathrm{Fe}, \mathrm{S}$ and and O respectively are 56, 32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]

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