Question

The molecular weight of a gas that has a density of 7.10 g/L at 25.0 C and 1.00 atm pressure is g/mol.
A) 6.85× 10-2
B) 14.6
C) 28.0
D) 174

Answer 1

Answer:

D) 174 g/mol

Explanation:

Step 1: Given and required data

• Density of the gas (ρ): 7.10 g/L

• Temperature (T): 25.0 °C

• Pressure (P): 1.00 atm

• Ideal gas constant (R): 0.0821 atm.L/mol.K

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 25.0°C + 273.15

K = 298.2 K

Step 3: Calculate the molecular weight (M) of the gas

We will use the following expression derived from the ideal gas equation.

ρ = P × M / R × T

M = ρ × R × T / P

M = 7.10 g/L × (0.0821 atm.L/mol.K) × 298.2 K / 1.00 atm

M = 174 g/mol

Answer 2

The molecular weight of a gas will be "174".

Given values:

• Density, d = 7.10 g/L
• Temperature, T = 25.0°C
• Pressure, P = 1.00 atm

We know the formula,

→ $PV = nRT$

or,

→ $PM = \frac{W}{V} RT$

or,

→ $PM = dRT$

By substituting the values, we get

$1\times M = 7.10\times 0.0821\times 298$

     $M = 7.10\times 0.0821\times 298$

          $=174$

Thus the answer i.e., "option D" is appropriate.

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