
If the boiling point of water is ${100^o}C$. How much gram of $NaCl$ is added in $500g$ of water to increase its boiling point of water by approx. ${1^o}C$?
$[{({K_b})_{{H_2}O}} = 0.52K{\text{ kg mo}}{{\text{l}}^{ - 1}}]$
(A)- 2.812 g
(B)- 28.12g
(C)- 14.06g
(D)- 7.03g
Answer
569.7k+ views
Hint: Boiling-point elevation is a phenomenon which states that the boiling point of a liquid (a solvent) will be higher when another compound is added. In other words, a solution has a higher boiling point than a pure solvent. This happens whenever a non-volatile solute, such as a salt, is added to any pure solvent like water.
Complete step by step solution:
The extent of boiling-point elevation can be calculated by applying Clausius–Clapeyron equation and Raoult's law together by assuming that the solute which is being added in the solvent is non-volatile in nature. The result is that in dilute solutions, the extent of boiling-point elevation is directly proportional to the molal concentration of the solution according to the equation:
$\Delta {T_b} = i \times {K_b} \times m$
where $\Delta {T_b}$ is the elevation in boiling point of the liquid.
$i$ is the van't Hoff factor
${K_b}$ is the ebullioscopic constant, which is dependent on the properties of the solvent.
and m is the molality of the given solution.
At high concentration, the above formula is less precise due to nonideality of the solution. If the solute is also volatile, one of the key assumptions used while deriving the formula will not be true, since it is derived for the solutions of non-volatile solutes in a volatile solvent.
Now, putting the respective terms given in the question in the above formula, we will get
$\begin{align}
&\implies 1 = 2 \times 0.52 \times \dfrac{w}{{58.5}} \times \dfrac{{1000}}{{500}} \\
&\therefore w = 28.12g \\
\end{align} $
As we can see, 28.12 gram of NaCl should be added in 500g of water to increase its boiling point by approx. ${1^o}C$.
Therefore, the correct answer of the given question is option (B)- 28.12g.
Note: Always remember that the boiling point elevation is a colligative property, which means that it is dependent on the presence of dissolved particles and their number in the solution, but not in the identity of the given solute.
Complete step by step solution:
The extent of boiling-point elevation can be calculated by applying Clausius–Clapeyron equation and Raoult's law together by assuming that the solute which is being added in the solvent is non-volatile in nature. The result is that in dilute solutions, the extent of boiling-point elevation is directly proportional to the molal concentration of the solution according to the equation:
$\Delta {T_b} = i \times {K_b} \times m$
where $\Delta {T_b}$ is the elevation in boiling point of the liquid.
$i$ is the van't Hoff factor
${K_b}$ is the ebullioscopic constant, which is dependent on the properties of the solvent.
and m is the molality of the given solution.
At high concentration, the above formula is less precise due to nonideality of the solution. If the solute is also volatile, one of the key assumptions used while deriving the formula will not be true, since it is derived for the solutions of non-volatile solutes in a volatile solvent.
Now, putting the respective terms given in the question in the above formula, we will get
$\begin{align}
&\implies 1 = 2 \times 0.52 \times \dfrac{w}{{58.5}} \times \dfrac{{1000}}{{500}} \\
&\therefore w = 28.12g \\
\end{align} $
As we can see, 28.12 gram of NaCl should be added in 500g of water to increase its boiling point by approx. ${1^o}C$.
Therefore, the correct answer of the given question is option (B)- 28.12g.
Note: Always remember that the boiling point elevation is a colligative property, which means that it is dependent on the presence of dissolved particles and their number in the solution, but not in the identity of the given solute.
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