Question

Calculate the volume occupied by 0,5mol of sulphur dioxide at STP​

Answer 1

Hi,

To calculate the volume occupied by 0.5 mol of sulfur dioxide (SO2) at standard temperature and pressure (STP), we can use the ideal gas law equation, PV = nRT. STP conditions are defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (atm).

1. Convert the temperature from Celsius to Kelvin:

STP temperature (T) = 0 °C + 273.15 = 273.15 K

2. Determine the value of the ideal gas constant (R):

The ideal gas constant (R) is equal to 0.0821 L·atm/(mol·K).

3. Plug in the values into the ideal gas law equation:

PV = nRT

P = 1 atm (given)

V = volume (to be calculated)

n = 0.5 mol (given)

R = 0.0821 L·atm/(mol·K)

T = 273.15 K (STP temperature)

Solving for V:

V = (nRT) / P

V = (0.5 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)

4. Calculate the volume:

V = 11.2 L

Therefore, 0.5 mol of sulfur dioxide occupies a volume of 11.2 liters at STP.

I hope that this helped you. :)

Answer 2

0.5 mol of sulfur dioxide occupies approximately 11.18875 liters of volume at STP.

Step-by-step explanation:

To calculate the volume occupied by 0.5 mol of sulfur dioxide (SO₂) at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:

PV = nRT

Where:

- P is the pressure (at STP, it is 1 atm)

- V is the volume

- n is the number of moles

- R is the ideal gas constant (0.0821 L·atm/(mol·K))

- T is the temperature in Kelvin (at STP, it is 273 K)

Given that we have 0.5 mol of SO₂, we can substitute the values into the equation and solve for V:

(1 atm) * V = (0.5 mol) * (0.0821 L·atm/(mol·K)) * (273 K)

Simplifying the equation:

V = (0.5 mol) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm)

V = 11.18875 L

Therefore, 0.5 mol of sulfur dioxide occupies approximately 11.18875 liters of volume at STP.