The Question is from neet-chemistry solutions abnormal-molar-mass, and it was created by the author A.RAMA NAGESWARARAO

Question

A sample of a volatile liquid was vaporized in a Victor Meyer's apparatus at 100°C and 750 mmHg pressure. The volume of air collected over water at 25°C was 30 mL.  The atmospheric pressure is 760 mmHg and the aqueous tension at 25°C is 24 mmHg. If the mass of the liquid taken was 0.06 g, what is the approximate molar mass of the liquid?

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Ideal Gas Law and Vapor Density
 **Related Topics**: Molar mass determination, Victor Meyer's method
 **Prerequisites**: Understanding of ideal gas equation ($PV = nRT$), Dalton's law of partial pressures, unit conversions.
## SOLUTION PATHWAY
**Given Information**:
  $V_{air} = 30 mL = 30 	imes 10^{-3} L$
  $T_{air} = 25°C = 298 K$
  $P_{atm} = 760 mmHg$
  $P_{water} = 24 mmHg$
  $T_{liquid} = 100°C = 373 K$
  $P_{liquid} = 750 mmHg$
  $m_{liquid} = 0.06 g$
  $R = 0.0821 L atm mol⁻¹ K⁻¹ (Ideal gas constant)
**Method & Steps**:
 1. **Calculate the pressure of dry air collected:**
  Reasoning:  The air collected is saturated with water vapor. We need to find the partial pressure of dry air using Dalton's Law.
  Formula/Rule: $P_{dry air} = P_{atm} - P_{water} = 760 mmHg - 24 mmHg = 736 mmHg$
  Convert to atm: $P_{dry air} = 736 mmHg 	imes \frac{1 atm}{760 mmHg} = 0.9684 atm$
 2. **Use the Ideal Gas Law to find the number of moles of dry air:**
  Reasoning: The volume of dry air collected is equal to the volume of vapor produced by the liquid.
  Formula/Rule: $PV = nRT$
  $n = \frac{PV}{RT} = \frac{0.9684 atm 	imes 30 	imes 10^{-3} L}{0.0821 L atm mol⁻¹ K⁻¹ 	imes 298 K} = 0.00118 mol$
 3. **Apply the Ideal Gas Law to the liquid vapor at 100°C and 750 mmHg:**
  Reasoning: The number of moles of dry air collected equals the number of moles of the vaporized liquid (assuming ideal behavior).
  Formula/Rule: $PV = nRT$, but we use the conditions of the liquid vapor.
  Convert pressure to atm: $P_{liquid} = 750 mmHg 	imes \frac{1 atm}{760 mmHg} = 0.9868 atm$
  Since the number of moles is the same:  $n = \frac{P_{liquid}V_{liquid}}{RT_{liquid}}$, where $V_{liquid}$ is unknown, but we know $n = 0.00118 mol$.
 4. **Calculate the molar mass of the liquid:**
  Reasoning: Molar mass (M) = mass (m) / moles (n)
  $M = \frac{0.06 g}{0.00118 mol} \approx 50.8 g/mol$
  Result: The approximate molar mass is 50.8 g/mol. This is closest to option C (60 g/mol).
  Verification: The calculation involves several approximations (ideal gas behavior, etc.), so a small discrepancy is expected.
## OPTION ANALYSIS
A :x: Incorrect calculation or misunderstanding of the concept.
B :x: Incorrect calculation or misunderstanding of the concept.
C ✓: Closest to the calculated molar mass (50.8 g/mol).
D :x: Incorrect calculation or misunderstanding of the concept.
## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the aqueous tension when calculating the partial pressure of dry air.: Correctly subtract the aqueous tension from the atmospheric pressure.
  2.Using incorrect units or failing to convert units consistently.: Ensure consistent use of units (atm, L, mol, K).
 **Quick Tips**:
  1.Remember to always account for water vapor pressure when collecting gases over water.
  2.Double-check your unit conversions before applying the ideal gas law.

A.RAMA NAGESWARARAO · Answer

30 g/mol

A.RAMA NAGESWARARAO · Answer

45 g/mol

A.RAMA NAGESWARARAO · Answer

The correct Option is "60 g/mol"

A.RAMA NAGESWARARAO · Answer

75 g/mol

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