The correct answer is 40
Mass of organic compound = 0.25 g
Mass of AgCl = 0.40 g
%\(Cl=\frac{35.5×mass\ of AgCl}{143.5×(mass\ of\ organic\ compound)}×100\)
\(=\frac{35.5×0.40×100}{143.5×0.25 }\)
= 39.581 ≃ 40%
Therefore , Cl = 40 %
List-I (Name of the test) | List-II (Reaction sequence involved) [M is metal] | |
---|---|---|
(A) Borax bead test | I | ![]() |
(B) Charcoal cavity test | II | ![]() |
(C) Cobalt nitrate test | III | ![]() |
(D) Flame test | IV | ![]() |
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32