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How many grams of fluorine gas occupy 543.2ml and exert 123456 pa of pressure at 29.8 degrees Celsius

User Vnk
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1 Answer

28 votes
28 votes

Answer:

m = 506 grams of fluorine

Step-by-step explanation:

Given:

M = 19·10⁻³ kg / mol - molar mass of fluorine

V = 543.2 mL = 543.2·10⁻³ m³

p = 123456 Pa

t = 29.8°C

_____________________

m - ?

Temperature in degrees Kelvin:

T = 273.15 + t = 273.15 + 29.8 = 302.95 K

Clapeyron-Mendeleev law:

p·V = (m/M)·R·T

Mass of fluorine:

m = p·V·M / (R·T)

m = 123456·543.2·10⁻³·19·10⁻³ / (8,31·302.95) ≈ 0.506 kg

or m = 506 g

User Tafaust
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