1
JEE Advanced 2025 Paper 2 Online
Numerical
+4
-0
Change Language

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is $\frac{\boldsymbol{X}}{F} \times 10^3$ volts, where $F$ is the Faraday constant. The value of $\boldsymbol{X}$ is _____________.

Use: Standard Gibbs energies of formation at 298 K are: $\Delta_f G_{\mathrm{CO}_2}^o=-394 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {water }}^o=$ $-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^o=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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2
JEE Advanced 2025 Paper 2 Online
Numerical
+4
-0
Change Language
The sum of the spin only magnetic moment values (in B.M.) of $\left[\mathrm{Mn}(\mathrm{Br})_6\right]^{3-}$ and $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$ is _________.
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3
JEE Advanced 2025 Paper 2 Online
Numerical
+4
-0
Change Language

A linear octasaccharide (molar mass $=1024 \mathrm{~g} \mathrm{~mol}^{-1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is $58.26 \%(\mathrm{w} / \mathrm{w})$ of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is $\qquad$ .

Use: Molar mass $\left(\right.$ in g $\left.\mathrm{mol}^{-1}\right)$ : ribose $=150,2$-deoxyribose $=134$, glucose $=180$;

Atomic mass (in amu): $\mathrm{H}=1, \mathrm{O}=16$

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4
JEE Advanced 2025 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1
Change Language

Let $x_0$ be the real number such that $e^{x_0} + x_0 = 0$. For a given real number $\alpha$, define

$$g(x) = \frac{3x e^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$$

for all real numbers $x$.

Then which one of the following statements is TRUE?

A

For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

B

For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1$

C

For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

D

For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}$

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