Question

When ${\text{4}}{\text{.8}}$gm of substance is dissolve in ${\text{240}}$gm of water, then the solution obtain boils at ${\text{100}}{\text{.7}}\,{\,^{\text{o}}}{\text{C}}$ . Determine the molecular weight of substance (given ${{\text{K}}_{\text{b}}}$= ${\text{0}}{\text{.52}}$)

Answer 1

Hint:To answer this question we should know the formulas of elevation in boiling point, molality and mole. First we will determine the molality by using elevation in boiling point formula. Then by using molality formula we will determine the moles of solute. Then by using mole formula we can determine the molecular mass of solute.


Complete solution:
The formula of elevation in boiling point is as follows:
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = }}{{\text{K}}_{\text{b}}}{\text{.}}\,{\text{m}}$
Where,
$\delta {T_b}\,$is the elevation in boiling point.
${K_b}$is the boiling point elevation constant.
We will determine the elevation in boiling point as follows:
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 ${\text{100}}{\text{.7}}\,{\,^{\text{o}}}{\text{C}}$ for boiling point of solution.
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = }}\,{\text{100}}{\text{.7}}\,{\,^{\text{o}}}{\text{C}}\, - {\text{100}}\,{\,^{\text{o}}}{\text{C}}$
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = }}\,{\text{0}}{\text{.7}}\,{\,^{\text{o}}}{\text{C}}\,$
We will rearrange the boiling point elevation formula for molality as follows:
${\delta }{{\text{T}}_{\text{b}}}\,{\text{ = }}{{\text{K}}_{\text{b}}}{\text{.}}\,{\text{m}}$
\[{\text{m}}\,{\text{ = }}\,\dfrac{{{\delta }{{\text{T}}_{\text{b}}}}}{{{{\text{K}}_{\text{b}}}}}\]
On substituting $0.52$ for boiling point elevation constant and \[0.7\]for elevation in boiling point.
\[{\text{m}}\, = \,\dfrac{{0.7}}{{0.52}}\]
\[{\text{m}}\, = \,1.35\,\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}\]
So, the molality of the solution is\[1.35\,\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}\].
We will convert the amount of water from gram to kilogram as follows:
$1000{\text{g = }}\,{\text{1}}\,{\text{kg}}$
$240{\text{g = }}\,0.240\,{\text{kg}}$
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 $0.240$kg for mass of solvent and \[1.35\,\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}\] for molality.
$1.35\,\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}\,{\text{ = }}\dfrac{{{\text{mole}}\,{\text{of}}\,{\text{solute}}}}{{{\text{0}}{\text{.240}}\,{\text{kg}}}}$
${\text{mole}}\,{\text{of}}\,{\text{solute}} = 1.35\,\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}} \times 0.240\,{\text{kg}}\,$
${\text{mole}}\,{\text{of}}\,{\text{solute}} = 0.3230\,{\text{mol}}\,\,$
So, the mole of solute is$0.3230\,{\text{mol}}\,$.
We will use the mole formula to determine the molar mass of unknown material.
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
The mass of solute is $4.8$gram.
On substituting $0.3230\,{\text{mol}}\,$for mole and $4.8$g for mass.
\[0.3230\,{\text{mol}}\,\,\,{\text{ = }}\,\dfrac{{4.8\,{\text{g}}}}{{{\text{molar }}\,{\text{mass}}\,}}\]
\[{\text{molar }}\,{\text{mass}}\,{\text{ = }}\,\dfrac{{4.8\,{\text{g}}}}{{0.3230\,{\text{mol}}\,}}\]
\[{\text{molar mass}} = \,{\text{14}}{\text{.86 g /mol}}\]

Therefore, the molar mass of the solute is\[{\text{14}}{\text{.86}}\] g/mol.


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.