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A 353.2mL sample of chlorine gas is collected at 25.2°C and an atmospheric pressure of 100.8kPa What would the volume be at STP?

User Rpitting
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Answer:

Step-by-step explanation:

To solve this problem, we can use the ideal gas law:

PV = nRT

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

At STP (standard temperature and pressure), the temperature is 0°C or 273.15 K, and the pressure is 1 atm or 101.325 kPa.

We can use the ideal gas law to find the number of moles of chlorine gas in the sample:

n = PV/RT

where P is the pressure, V is the volume, R is the gas constant, and T is the temperature in Kelvin.

First, we need to convert the given temperature of 25.2°C to Kelvin:

T = 25.2°C + 273.15 = 298.35 K

Now we can calculate the number of moles of chlorine gas in the sample:

n = (100.8 kPa)(353.2 mL)/(8.314 J/K/mol)(298.35 K)

n = 0.0158 mol

Next, we can use the number of moles and the ideal gas law to find the volume at STP:

V = nRT/P

V = (0.0158 mol)(8.314 J/K/mol)(273.15 K)/(101.325 kPa)

V = 0.364 L or 364 mL

Therefore, the volume of the chlorine gas at STP would be 364 mL.

User Stevemo
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