Question

The elevation in boiling point of a solution of $13.44\,{\text{g}}$ of ${\text{CuC}}{{\text{l}}_2}$, (molecular weight$= 134.4,$ ${{\text{K}}_{\text{b}}}{\text{ = }}\,0.52\,{\text{K}}\,{\text{Molality}}^{-1}$) in 1 kg water using the following information will be
A) 0.16
B) 0.05
C) 0.1
D) 0.2

Answer 1

Elevation in boiling point refers to increase in the boiling point of solute upon addition of the solute. The elevation in boiling point is the product of the boiling point elevation constant and molality.

Formula used: $\Delta {T_b}\, = i\,{K_b}.\,m$

Complete step by step answer:
The formula of freezing point depression is as follows:
$\Delta {T_b}\, = i\,{K_b}.\,m$
Where,
$\Delta {T_b}$ is the elevation in boiling point.
${K_b}$ is the boiling point elevation constant.
$i\,$ is the van't Hoff factor.
Determine the van't Hoff factor for Copper (II) chloride as follows:
Copper (II) chloride ${\text{CuC}}{{\text{l}}_2}$ is an ionic compound which dissociates in water as follows:
${\text{CuC}}{{\text{l}}_2}\,\mathop \to \limits^{{{\text{H}}_{\text{2}}}{\text{O}}} \,{\text{C}}{{\text{u}}^{{\text{2 + }}}}\, + \,{\text{2 C}}{{\text{l}}^ - }$
Copper (II) chloride produces three ions so the value of van’t Hoff factor is 3.
Determine the mole of Copper (II) chloride as follows:
The formula of the mole is as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Given mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
The molar mass of Copper (II) chloride is $134.4\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,$.
Substitute $134.4\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,$ for molar mass and $13.44\,{\text{g}}$ for mass of copper (II) chloride.
${\text{Mole}}\,{\text{ = }}\,\dfrac{{13.44\,{\text{g}}}}{{134.4\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}$
${\text{Mole}}\,{\text{ = }}\,0.099\,{\text{mol}}$
Determine the molality of Copper (II) chloride as follows:
Molality is defined as the amount of solute (Copper (II) chloride) dissolved in an amount of solvent (water).
The formula of molality is as follows:
${\text{Molality}}\,{\text{ = }}\,\dfrac{{{\text{Mole of solute}}}}{{{\text{kg}}\,{\text{of}}\,{\text{solvent}}}}$
Substitute $0.099\,{\text{mol}}$ for mole of solute and $1\,{\text{kg}}$ for kg of solvent.
${\text{Molality}}\,{\text{ = }}\,\dfrac{{0.099\,{\text{mol}}}}{{{\text{1}}\,{\text{kg}}}}$
${\text{Molality}}\,{\text{ = }}\,0.099\,{\text{m}}$
Determine the elevation in boiling point as follows:
Substitute 3 for van't Hoff factor, $0.52\,{\text{K}}\,/{\text{Molality}}$ for boiling point elevation constant and $0.099\,{\text{m}}$ for molality.
$\Delta {T_b}\, = 3\, \times 0.52\,{\text{K}}\,/{\text{Molality}}\, \times 0.099\,{\text{m}}$
$\Delta {T_b}\, = 0.16\,{\text{K}}$
So, the elevation in boiling point is $0.16\,{\text{K}}$.

Therefore, option (A) 0.16, is correct.

Note: When a solute is added to the pure solvent, the boiling point of the solution increases which is known as the elevation in boiling point. The boiling point of the pure solvent at a temperature is defined as the boiling point constant. The Van't Hoff factor represents the degree of dissociation or number of ions produced by a compound on dissolution.