Question

Solution of $1$ mol NaCl in ${\text{1000}}$g ${{\text{H}}_{\text{2}}}{\text{O}}$ will change the boiling point of water to.
A. ${\text{100}}{\text{.51}}{\,^{\text{o}}}{\text{C}}$
B. ${\text{101}}{\text{.02}}{\,^{\text{o}}}{\text{C}}$
C. ${\text{101}}{\text{.53}}{\,^{\text{o}}}{\text{C}}$
D. ${\text{101}}{\text{.86}}{\,^{\text{o}}}{\text{C}}$
E. ${\text{103}}{\text{.6}}{\,^{\text{o}}}{\text{C}}$

Answer 1

To answer this question we should know the formulas of elevation in boiling point, and molality. First we will determine the molality by using molality formula. Then by using boiling point elevation formula we will determine the increase in temperature. Then we add the increases in temperature to the boiling point of water to get the change in boiling point of water.

Complete solution:
First, we will convert the mass of solvent (water) form gram to kilogram as follows:
$1000{\text{g = }}\,{\text{1}}\,{\text{kg}}$
The formula of molarity is as follows:
${\text{molality}}\,{\text{ = }}\dfrac{{{\text{mole}}\,{\text{of}}\,{\text{solute}}}}{{{\text{kg}}\,{\text{of}}\,{\text{solvent}}}}$
On substituting $1$kg for mass of solvent and $1$ mol for mole of solute.
${\text{molality}}\,{\text{ = }}\dfrac{{{\text{1}}\,{\text{mol}}}}{{{\text{1}}\,{\text{kg}}}}$
${\text{molality}}\,{\text{ = 1}}\,{\text{m}}$
So, the molality of NaCl solution is${\text{1}}$m.
Now, we will determine the van’t Hoff factor for NaCl as follows:
${\text{NaCl}}\,\, \to \,{\text{N}}{{\text{a}}^{ + \,}} + \,\,{\text{C}}{{\text{l}}^ - }$
Sodium chloride produces two ions so the van’t Hoff factor for NaCl is $2$ .
The formula of elevation in boiling point is as follows:
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = i}}{{\text{K}}_{\text{b}}}{\text{.}}\,{\text{m}}$
Where,
${\delta }{{\text{T}}_{\text{b}}}\,$is the elevation in boiling point.
${\text{i}}$ is the van’t Hoff factor.
${{\text{K}}_{\text{b}}}$is the boiling point elevation constant.
We will determine the elevation in boiling point as follows:
On substituting $2$ for i, ${\text{0}}{\text{.512}}{\,^{\text{o}}}{\text{C}}\,{{\text{m}}^{ - 1}}$ for ${{\text{K}}_{\text{b}}}$ and $1$for m.
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = i}}{{\text{K}}_{\text{b}}}{\text{.}}\,{\text{m}}$
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = 2}} \times \,{\text{0}}{\text{.512}}{\,^{\text{o}}}{\text{C}}\,{{\text{m}}^{ - 1}} \times \,1{\text{m}}$
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = }}1.02{\,^{\text{o}}}{\text{C}}$
So, the elevation in boiling point is $1.02{\,^{\text{o}}}{\text{C}}$.
Now, the boiling point of NaCl solution is,
The boiling point of pure water is${100^{\text{o}}}{\text{C}}$.
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = }}\,{{\text{T}}_{{\text{solution}}}}\, - {{\text{T}}_{{\text{water}}}}$
On substituting ${100^{\text{o}}}{\text{C}}$for boiling point of pure water and $1.02{\,^{\text{o}}}{\text{C}}$ for elevation in boiling point.
$1.02{\,^{\text{o}}}{\text{C}}\,{\text{ = }}\,{{\text{T}}_{{\text{solution}}}}\, - {\text{100}}\,{\,^{\text{o}}}{\text{C}}$
$\,{{\text{T}}_{{\text{solution}}}}\, = \,\,1.02{\,^{\text{o}}}{\text{C}}\, + {\text{100}}\,{\,^{\text{o}}}{\text{C}}$
$\,{{\text{T}}_{{\text{solution}}}}\, = \,{\text{101}}{\text{.02}}\,{\,^{\text{o}}}{\text{C}}$
So, $1$ mol NaCl in ${\text{1000}}$g ${{\text{H}}_{\text{2}}}{\text{O}}$ will change the boiling point of water to ${\text{101}}{\text{.02}}{\,^{\text{o}}}{\text{C}}$.

Therefore, option (B) ${\text{101}}{\text{.02}}{\,^{\text{o}}}{\text{C}}$ is correct.

Note: The elevation in boiling point is the product of the boiling point elevation constant and molality. Molality is defined as the mole of solute dissolved in a kilogram of solvent (water). Here, unit is important. When a solute is added to the pure solvent, the boiling point of the solution increases which is known as the elevation in the boiling point. So, the boiling point of the solution will be more than the boiling point of the solvent. The boiling point of the pure solvent at a temperature is defined as the boiling point constant.