Answer to Question #340516 in General Chemistry for aqll

Question

What is the Volume of the water vapour if 3.2 grams of water vapour is observed at 22 C

and 125 kPa. ?    R = 8.31 L kPa/ K mol 

Answer 1

Given:

Mass of water is 3.2 g

Molar mass of water is 18.0153 g mol⁻¹ 

T = 22°C + 273.15 = 295.15 K

P = 125 kPa

R = 8.31 L kPa K⁻¹ mol⁻¹ 

Solution:

Moles = Mass / Molar mass

Therefore,

Moles of water = (3.2 g) / (18.0153 g mol⁻¹) = 0.1776 mol

n = 0.1776 mol

The ideal gas law relates the amount of gas present to its pressure, volume, and temperature.

The ideal gas law is written as PV = nRT, 

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

 

Rearrange the equation to solve for V:

V = nRT / P

Plug in the numbers:

V = (0.1776 mol × 8.31 L kPa K⁻¹ mol⁻¹ × 295.15 K) / (125 kPa) = 3.4848 L = 3.5 L

V = 3.5 L

Answer: The volume of the water vapour is 3.5 liters