1
JEE Main 2025 (Online) 8th April Evening Shift
Numerical
+4
-1

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _______ M. (Nearest Integer value)

(Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol-1)

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2
JEE Main 2025 (Online) 8th April Evening Shift
Numerical
+4
-1

Resonance in $\mathrm{X}_2 \mathrm{Y}$ can be represented as

JEE Main 2025 (Online) 8th April Evening Shift Chemistry - Thermodynamics Question 6 English

The enthalpy of formation of $X_2Y$ $ \left(X = X(g) + \frac{1}{2} Y = Y(g) \rightarrow X_2Y(g) \right) $ is 80 kJ mol$^{-1}$. The magnitude of resonance energy of $X_2Y$ is __ kJ mol$^{-1}$ (nearest integer value).

Given: Bond energies of $X \equiv X$, $X = X$, $Y = Y$ and $X = Y$ are 940, 410, 500, and 602 kJ mol$^{-1}$ respectively.
valence $X$: 3, $Y$: 2

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3
JEE Main 2025 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the
points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

A

$3$

B

$\frac{5\sqrt{2}}{3}$

C

$\frac{\sqrt{2}}{3}$

D

$4$

4
JEE Main 2025 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f(f(x))\right) - \frac{y}{\sqrt{x}} \right)$, $y(0) = 0$, then $y(1)$ is

A

$\frac{1 - e^3}{e^4}$

B

$\frac{e-1}{e^4}$

C

$\frac{1 - e^2}{e^4}$

D

$\frac{2e - 1}{e^3}$

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