Question

(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass${\text{ = 180}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$ ) in water is labeled as $10\% $ (by mass). What would be the molality and molarity of the solution? (density of water$ = 1.2\,{\text{g}}\,{\text{. m}}{{\text{L}}^{ - 1}}$ )

Answer 1

Hint: When the solute and solvent shows weaker interaction in solution than the interaction of pure solute-solute and solvent –solvent, they deviated positively from Raoult’s law. When the solute and solvent shows stronger interaction in solution than the interaction of pure solute-solute and solvent –solvent, they deviated negatively from Raoult’s law.

Formula used: ${\text{Molarity}}\,\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{solution}}}}$
${\text{Molality}}\,\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{kg}}\,{\text{of}}\,{\text{solvent}}}}$

Complete step by step answer:
(a) According to Raoult’s law, the partial pressure of a component is the product of vapour pressure of pure solvent and mole fraction of that component.
When a solution shows deviation from Raoult’s law over the complete range of concentration, the solution is known as a non-ideal solution.
The vapour pressure of the non-ideal solution can be higher or lower than the vapour pressure predicted by Raoult’s law.
If the vapour pressure of the non-ideal solution is higher than the vapour pressure predicted by Raoult’s law, the deviation is known as positive deviation.
If the vapour pressure of the non-ideal solution is lower than the vapour pressure predicted by Raoult’s law, the deviation is known as negative deviation.
The reason for the deviation is molecular interactions.
${\text{A}} - {\text{B}}\,\,{\text{ < }}\,{\text{A}} - {\text{A}}\,{\text{, B}} - {\text{B}}$ for positive deviation
Here, A is the solute and B is the solvent so, A-B shows the interaction between solute and solvent.
for negative deviation ${\text{A}} - {\text{B}}\,\,{\text{ > }}\,{\text{A}} - {\text{A}}\,{\text{, B}} - {\text{B}}$
The type of deviation shown by a mixture of acetone and ethanol is explained as follows:
Ethanol has hydrogen bonding in pure form. Hydrogen bonding provides extra stability. When acetone is added to ethanol, acetone breaks some of the hydrogen bonds of ethanol thus, decreasing the interaction. So, the ethanol-acetone shows weaker interaction than pure ethanol-ethanol interaction.
Therefore, the mixture of ethanol and acetone shows a positive deviation from Raoult’s law.
(b) $10\% $ (by mass) means $10\,{\text{g}}$ glucose is dissolved in $100\,{\text{g}}$of solution.
So, the amount of solvent (water) is $90\,{\text{g}}$.
Determine the mole of glucose as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{\,{\text{Molar}}\,{\text{mass}}}}$
Substitute 10 g for mass and 180 g/mol for molar mass.
${\text{Mole}}\,{\text{ = }}\,{\dfrac{10\,g}{180\,g/mol}}$
${\text{mole}}\,{\text{ = }}\,0.05\,{\text{mol}}$
So, the mole of glucose is $0.05\,{\text{mol}}$.
Convert the amount of water from g to kg as follows:
${\text{1000}}\,{\text{g = }}\,{\text{1}}\,{\text{kg}}$
$90\,{\text{g}}\,{\text{ = }}\,0.090\,{\text{kg}}$
Substitute $0.05\,{\text{mol}}$for mole of solute (glucose) and $0.090\,{\text{kg}}$ for mass of solvent in molality formula.
${\text{Molality}}\,{\text{ = }}\,\dfrac{{0.05\,{\text{mol}}}}{{\,0.090\,{\text{kg}}}}$
${\text{Molality}}\,{\text{ = }}\,0.617\,{\text{m}}$
So, the molality is $0.6\,{\text{m}}$.
Determine the volume of water as follows:
${\text{Density}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}$
Substitute $1.2\,{\text{g}}\,{\text{. m}}{{\text{L}}^{ - 1}}$ for density and ${\text{100g}}$ for mass.
$1.2\,{\text{g}}\,{\text{. m}}{{\text{L}}^{ - 1}}\,{\text{ = }}\,\dfrac{{100\,{\text{g}}}}{{{\text{Volume}}}}$
${\text{Volume}}\,{\text{ = }}\,\dfrac{{100\,{\text{g}}}}{{1.2\,{\text{g}}\,{\text{. m}}{{\text{L}}^{ - 1}}}}$
${\text{Volume}}\,{\text{ = }}\,83.3\,{\text{mL}}$
Convert the volume of water from mL to L as follows:
$1000\,{\text{mL}}\,{\text{ = }}\,{\text{1 L}}$
$83.3\,{\text{mL}} = 0.083\,{\text{L}}$
Determine the molarity as follows:
Substitute $0.083\,{\text{L}}$ for volume of solution and $0.05\,{\text{mol}}$ for mole of glucose.
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{0.05\,{\text{mol}}}}{{0.083\,\,{\text{L}}}}$
${\text{Molarity}}\,{\text{ = }}\,{\text{0}}{\text{.6}}\,{\text{M}}$
So, the Molarity is ${\text{0}}{\text{.6}}\,{\text{M}}$.

Therefore, the molarity of the solution is ${\text{0}}{\text{.6}}\,{\text{M}}$ and the molality of the solution is $0.6\,{\text{m}}$.

Note: Molarity is defined as the mole of solute dissolved in per litter of the solution and molality is defined as the mole of solute dissolved in per kg of the solvent. Both are the concentration terms.