\(\frac{1}{2} \times 1.5 \times 60^2 \times \frac{1}{4} = 100 \times 0.42 \times \triangle \;T\)
\(\triangle\; T = \frac{1.5\times 60^2}{8\times 100 \times 0.42} = 16.07 \;\degree C\)


A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.
The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,
W = ΔKE = ½ mv2f − ½ mv2i
Where,
vi → Speeds of the particle before the application of force
vf → Speeds of the particle after the application of force
m → Particle’s mass
Note: Energy and Momentum are related by, E = p2 / 2m.