Question

What would be the molality of \(1.5\,g\) of ethanoic acid (\(CH_3COOH\)) in \(75\,g\) of benzene?

• A. \(0.55\ mol\,kg^{-1}\)
• B. \(0.33\ mol\,kg^{-1}\)
• C. \(0.44\ mol\,kg^{-1}\)
• D. \(0.22\ mol\,kg^{-1}\)
• A. It must obey Raoult's law.
• B. \(\Delta_{mix}H=0\)
• C. \(\Delta_{mix}V=0\)
• D. \(\Delta_{mix}H\neq0\), and \(\Delta_{mix}V=0\)
• A. \(170\,g\,mol^{-1}\)
• B. \(190\,g\,mol^{-1}\)
• C. \(210\,g\,mol^{-1}\)
• D. \(180\,g\,mol^{-1}\)
• A. shows positive deviation from Raoult's law.
• B. shows negative deviation from Raoult's law.
• C. obeys Raoult's law.
• D. shows both negative and positive deviation from Raoult's law.
• A. Molality
• B. Mole fraction
• C. Molarity
• D. ppm

Answer 1

The Correct Option is B

Concept: Molality is defined as the number of moles of solute present in one kilogram of solvent. \[ m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \]

Step 1: Calculate molar mass of ethanoic acid.
\[ CH_3COOH=C_2H_4O_2 \] \[ =2(12)+4(1)+2(16) \] \[ =24+4+32 \] \[ =60\ g\,mol^{-1} \]

Step 2: Calculate moles of ethanoic acid.
\[ \text{Moles} = \frac{1.5}{60} \] \[ =0.025 \]

Step 3: Convert solvent mass into kilogram.
\[ 75g=0.075kg \]

Step 4: Calculate molality.
\[ m = \frac{0.025}{0.075} \] \[ =0.333 \] \[ \approx0.33\ mol\,kg^{-1} \]

Step 5: Final answer.
\[ \boxed{0.33\ mol\,kg^{-1}} \]

Answer 2

Concept: An ideal solution is one in which intermolecular forces between unlike molecules are nearly equal to those between like molecules. Because of this similarity in intermolecular interactions, mixing occurs without heat absorption, heat evolution or volume change.

Step 1: Recall the conditions for an ideal solution.
An ideal solution satisfies: \[ \Delta H_{mix}=0 \] and \[ \Delta V_{mix}=0 \] It also obeys Raoult's law throughout the entire concentration range.

Step 2: Check option (A).
Ideal solutions obey Raoult's law. Hence this statement is correct.

Step 3: Check option (B).
For ideal solutions, \[ \Delta H_{mix}=0 \] Hence correct.

Step 4: Check option (C).
For ideal solutions, \[ \Delta V_{mix}=0 \] Hence correct.

Step 5: Check option (D).
Option (D) states \[ \Delta H_{mix}\neq0 \] which contradicts the basic requirement of an ideal solution. Therefore it is incorrect. \[ \boxed{\Delta H_{mix}\neq0,\ \Delta V_{mix}=0} \] is not true for an ideal solution.

Answer 3

Concept: For a dilute solution containing a non-volatile solute, relative lowering of vapour pressure is equal to the mole fraction of solute. \[ \frac{P^0-P}{P^0}=X_B \] where \(P^0\) is vapour pressure of pure solvent and \(P\) is vapour pressure of solution.

Step 1: Calculate relative lowering of vapour pressure.
\[ P^0=0.950\;bar \] \[ P=0.945\;bar \] \[ \frac{P^0-P}{P^0} = \frac{0.950-0.945}{0.950} = \frac{0.005}{0.950} = 0.005263 \]

Step 2: Calculate moles of benzene.
\[ n_A=\frac{39}{78}=0.5\;mol \]

Step 3: Use mole fraction relation.
For dilute solution, \[ X_B=\frac{n_B}{n_A} \] \[ 0.005263=\frac{n_B}{0.5} \] \[ n_B=0.0026315\;mol \]

Step 4: Calculate molar mass of solute.
\[ M=\frac{\text{Mass}}{\text{Moles}} \] \[ M=\frac{0.5}{0.0026315} \] \[ M\approx190\,g\,mol^{-1} \]

Step 5: Final answer.
\[ \boxed{190\,g\,mol^{-1}} \]

Answer 4

Concept: Deviation from Raoult's law depends upon the relative strength of intermolecular forces between unlike molecules compared to those between like molecules. Positive deviation occurs when intermolecular attractions become weaker after mixing.

Step 1: Consider pure ethanol.
Ethanol molecules are strongly associated through intermolecular hydrogen bonding. \[ CH_3CH_2OH \cdots HOCH_2CH_3 \] These hydrogen bonds contribute significantly to molecular attraction.

Step 2: Mix acetone with ethanol.
Acetone molecules disrupt the extensive hydrogen bonding network present in pure ethanol. The new acetone-ethanol interactions are weaker than the ethanol-ethanol interactions.

Step 3: Effect on vapour pressure.
Because intermolecular attractions decrease upon mixing, molecules escape more easily into vapour phase. Hence vapour pressure becomes greater than predicted by Raoult's law. \[ P_{observed}>P_{Raoult} \]

Step 4: Final conclusion.
The solution shows positive deviation from Raoult's law. \[ \boxed{\text{Positive deviation}} \]

Answer 5

Concept: A concentration term becomes temperature dependent if it involves volume because volume changes with temperature.

Step 1: Examine molality.
\[ m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \] Mass does not change with temperature. Hence molality is temperature independent.

Step 2: Examine mole fraction.
\[ X=\frac{\text{Moles of component}}{\text{Total moles}} \] Since it depends only on moles, it is temperature independent.

Step 3: Examine ppm.
ppm generally depends on mass ratio. Therefore it is temperature independent.

Step 4: Examine molarity.
\[ M=\frac{\text{Moles of solute}}{\text{Volume of solution in litre}} \] Volume changes with temperature due to thermal expansion. Therefore molarity changes with temperature. \[ \boxed{\text{Molarity is temperature dependent}} \]