Question

What will be the boiling point of water if \[{{1 mole}}\] of ${{NaCl}}$ is dissolved in ${{1000 grams}}$ of water?
A. ${{100}}{{.51}}{{{ }}^{{o}}}{{C}}$
B. ${{101}}{{.02}}{{{ }}^{{o}}}{{C}}$
C. ${{101}}{{.53}}{{{ }}^{{o}}}{{C}}$
D. ${{101}}{{.86}}{{{ }}^{{o}}}{{C}}$
E. ${{103}}{{.62}}{{{ }}^{{o}}}{{C}}$

Answer 1

Hint: Molality is defined as the number of moles of solute per${{Kg}}$of solvent. Every liquid boils at a particular temperature where the vapour pressure of liquid becomes equal to atmospheric pressure and elevation in boiling point is the increase in the boiling point of the solution with the addition of a non- volatile solute.

Complete step by step answer:
Molality is defined as number of moles of solute per \[{{kg}}\] of solvent i.e.
${{Molality = }}\dfrac{{{{Moles\; of\; solute}}}}{{{{1\; kg\; of \;solvent}}}}$
In the above case ${{1mole}}$ of ${{NaCl}}$ is dissolved in ${{1000 grams}}$ of water
${{Molality = }}\dfrac{{{{Number\; of \;moles \;of \;NaCl}}}}{{{{Mass\; of \;water\; in kg}}}}$
\[{{Molality = }}\dfrac{{{{1 mol}}}}{{\dfrac{{{{1000 \times 1 kg}}}}{{{{1000}}}}}}{{ = 1 mol}}\]
According to elevation in boiling point
Elevation in boiling point ${{\Delta }}{{{T}}_{{b}}}$ is directly proportional to molal concentration of solute
${{\Delta }}{{{T}}_{{b}}}{{ \alpha m}}$
${{\Delta }}{{{T}}_{{b}}}{{ = i}}{{{K}}_{{b}}}{{m}}$ Where ${{{K}}_{{b}}}$ is boiling point elevation constant and value of ${{{K}}_{{b}}}$ for water is ${{0}}{{.51 Kkgmo}}{{{l}}^{{{ - 1}}}}$
i is Van’t Hoff factor and its value is 2 whereas 1molecule of ${{NaCl}}$ gives 2 ions of dissociation.
${{\Delta }}{{{T}}_{{b}}}{{ = 2 \times 0}}{{.51 \times 1 = 1}}{{.02}}{{{ }}^{{o}}}{{C}}$
Thus the boiling point of pure water is ${{100}}{{{ }}^{{o}}}{{C}}$
Thus for aqueous ${{NaCl}}$ boiling point of solution is
${{100 + 1}}{{.02 = 101}}{{.02}}{{{ }}^{{o}}}{{C}}$

So, the correct answer is B.

Additional information:
If in case \[{{{w}}_{{2}}}{{ gram}}\] of solute with molar mass ${{{M}}_{{2}}}$ is dissolved in ${{{w}}_{{1}}}{{ gram}}$ of solvent them ${{molality \;m}}$ of the solution is given by
${{m = }}\dfrac{{\dfrac{{{{{w}}_{{2}}}}}{{{{{M}}_{{2}}}}}}}{{\dfrac{{{{{w}}_{{1}}}}}{{{{1000}}}}}}{{ = }}\dfrac{{{{1000 \times }}{{{w}}_{{2}}}}}{{{{{M}}_{{2}}}{{ \times }}{{{w}}_{{1}}}}}$
On substituting the value of molality
${{\Delta }}{{{T}}_{{b}}}{{ = }}\dfrac{{{{{K}}_{{b}}}{{ \times 1000 \times }}{{{w}}_{{2}}}}}{{{{{M}}_{{2}}}{{ \times }}{{{w}}_{{1}}}}}$
${{{M}}_{{2}}}{{ = }}\dfrac{{{{1000 \times }}{{{w}}_{{2}}}{{ \times }}{{{K}}_{{b}}}}}{{{{\Delta }}{{{T}}_{{b}}}{{ \times }}{{{w}}_{{1}}}}}$$$$$

Note: The vapour pressure of solvent decreases in the presence of non-volatile solute. In such cases the solution should be boiled by applying a temperature greater than the boiling point of pure solvent such that its vapour pressure becomes equal to atmospheric pressure i.e. boiling point of solution will be always greater than boiling point of pure solvent thus the elevation in boiling point depends on the number of solvent molecules rather than their nature. When the solute is non-volatile only the solvent molecules contribute to the vapour pressure.
Elevation in boiling point is given as
${{\Delta }}{{{T}}_{{b}}}{{ = }}{{{T}}_{{b}}}{{ - }}{{{T}}_{{b}}}^{{o}}$ Where ${{{T}}_{{b}}}$the boiling is point of solution and ${{{T}}_{{b}}}^{{o}}$is the boiling point of solute.