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1
JEE Advanced 2025 Paper 1 Online
Numerical
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At 25 °C, the concentration of H+ ions in 1.00 × 10−3 M aqueous solution of a weak monobasic acid having acid dissociation constant (Ka) of 4.00 × 10−11 is X × 10−7 M. The value of X is ______.

Use: Ionic product of water (Kw) = 1.00 × 10−14 at 25 °C

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2
JEE Advanced 2025 Paper 1 Online
Numerical
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Molar volume (Vm) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with Vm as the variable. The ratio (in mol dm−3) of the coefficient of Vm2 to the coefficient of Vm for a gas having van der Waals constants a = 6.0 dm6 atm mol−2 and b = 0.060 dm3 mol−1 at 300 K and 300 atm is ______.

Use: Universal gas constant (R) = 0.082 dm3 atm mol−1 K−1

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3
JEE Advanced 2025 Paper 1 Online
Numerical
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Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is ______.

Use: Universal gas constant (R) = 8.3 J K−1 mol−1; Atomic mass (in amu): H = 1, O = 16

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4
JEE Advanced 2025 Paper 1 Online
Numerical
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The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ______.

Use: Atomic mass of N (in amu) = 14

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