Question

What is the expected boiling point elevation of water for a solution that contains $150g$ of sodium chloride dissolved in $1.0kg$ of water?

Answer 1

We have to know that the water bubbles at $100^\circ C$ at $1atm$ of pressing factor, yet an answer of saltwater does not. At the point when table salt is added to water, the subsequent arrangement has a higher edge of boiling over than the water did without anyone else.

Complete answer:
We need to realize that, we requested to track down the normal edge of boiling over rise of an answer given the measure of solute broke up in water.
By using the following equation,
$\Delta {T_b} = im{K_b}$
Where,
$i$ is the van’t Hoff factor,
$m$ is the molality,
$\Delta {T_b}$ is the boiling point elevation,
First, we have to calculate the number of moles,
No. of moles $= 150g{\text{NaCl}} \times \left( {\dfrac{{1mol{\text{NaCl}}}}{{58.44g{\text{NaCl}}}}} \right) = 2.57mol{\text{NaCl}}$
Then, we have to calculate the molality,
Molality =$\dfrac{{2.57mol}}{{1.0kg}} = 2.57m$
The boiling point constant for water = $0.512^\circ C/m$
The molal boiling point elevation constant for the solvent = ${K_b}$
Then, calculating the boiling point elevation water, by the following expression,
$\Delta {T_b} = 2 \times 2.57m \times 0.512^\circ C/m$
Hence, the answer is $2.6^\circ C$ .
When the boiling point of a fluid relies upon temperature, air pressure, and the fume pressing factor of the fluid. At the point when the environmental pressing factor is equivalent to the fume, pressing factor of the fluid, bubbling will start.

Note:
We have to see that the temperature is a proportion of the normal dynamic energy of the particles in issue. The Fahrenheit scale characterizes the edge of freezing over water as $32^\circ F$ and the limit as $212^\circ F$ . The Celsius scale sets the edge of freezing over and limit of water at $0^\circ C$ and $100^\circ C$ individually.