Question

A 0.239 gram sample of a gas in a 100-ml flask exerts a pressure of 603 mm Hg at 14 Celsius. What is the gas? (Hint: don’t forget BRINCLHOF)
A) chlorine

B) nitrogen

C) krypton

D) xenon

D) oxygen


Please help me and explain how u got ur answer

Answer 1

Answer:

A) chlorine

Explanation:

To solve this question we can use:

PV = nRT

In order to solve the moles of the gas. With the moles and the mass we can find the molar mass of the gas to have an idea of its identy:

PV = nRT

PV / RT = n

Where P is pressure: 603mmHg * (1atm / 760mmHg) = 0.7934atm

V = 100mL = 0.100L

R is gas constant = 0.082atmL/molK

T is absolute temperature = 14°C + 273.15 = 287.15K

0.7934atm*0.100L / 0.082atmL/molK*287.15K = n

3.37x10⁻³ moles of the gas

In 0.239g. The molar mass is:

0.239g / 3.37x10⁻³ moles = 70.9g/mol

The gas with this molar mass is Chlorine, Cl₂:

A) chlorine