Compare boiling point of given solutions:
(i) \( 10^{-4} \, \text{M NaCl} \)>(ii) \( 10^{-3} \, \text{M NaCl} \)>(iii) \( 10^{-2} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)
(iii) \( 10^{-2} \, \text{M NaCl} \)>(ii) \( 10^{-3} \, \text{M NaCl} \)>(i) \( 10^{-4} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)
(ii) \( 10^{-3} \, \text{M NaCl} \)>(i) \( 10^{-4} \, \text{M NaCl} \)>(iii) \( 10^{-2} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)
(iii) \( 10^{-2} \, \text{M NaCl} \)>(ii) \( 10^{-3} \, \text{M NaCl} \)>(i) \( 10^{-4} \, \text{M NaCl} \)>(iv) \( 10^{-4} \, \text{M urea} \)

Question

Let \(y=y(x)\) be the solution of the differential equation \[ x\frac{dy}{dx}=y-x^2\cot x,\quad x\in(0,\pi) \] If \(y\!\left(\frac{\pi}{2}\right)=\frac{\pi^2}{2}\), then \[ 6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right) \] is equal to:

Answer 1

A greater elevation in boiling point corresponds to a higher boiling point.

\( \Delta T_b \propto i \times m \)

For urea, \( i = 1 \).

For NaCl, \( i = 2 \).

Boiling point order: III>II>I>IV

Show Hint

The Correct Option is B

Solution and Explanation

Top Questions on Miscellaneous

Questions Asked in JEE Main exam