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# Answer to Question #162343 in Inorganic Chemistry for coim

Question #162343

a) Calculate the pressure exerted by 0.0153 moles of CO2 gas in a container of 1.467 L at 20.0C.

b) What is the volume of a 15.0 g sample of propane (C3H8) gas at Standard Temperature Pressure?

c) Assuming that the volume of the container in question part (b) above remains constant, what would be the pressure of the propane at room temperature (25 °C)?

d) A 1.00 g gaseous sample of hydrocarbon occupies a volume of 385 mL at 330 K and 1.00 atm. Find the molar mass of the compound.

e) Calculate the number of atoms of He(g) that are needed to exert a pressure of 4.0 X 10–2 kPa in a 15 mL container at a temperature of –23°C.

1
2021-02-10T01:34:18-0500

a) n=0.0153

V=1.467L

T=20+273 = 293K

Given the ideal gas equation

PV = nRT

"P=\\dfrac{nRT}{V}"

If we want P in atm,then R=0.08206"\\dfrac{L*atm}{mol*K}"

Therefore;

"P=\\dfrac{0.0153molxo0.08206L*atm\/mol*Kx293K}{1.467L}"

"P = 0.25076 atm"

b) STP conditions are;

T=273K

P=1atm

V=?

n=moles "=\\dfrac{mass}{MM}=\\dfrac{15.0g}{44.1g\/mol}=0.34mol"

PV = nRT

"V=\\dfrac{0.34molx0.08206Latm\/molKx273K}{1atm}" = 89.61L

c) Since V is constant we apply Gay Lussac's Law formula

"\\dfrac{P1}{T1}=\\dfrac{P2}{T2}"

P1=1atm

T1=273K

T2=298K

"P2=\\dfrac{1atmx298K}{273K}=1.09atm"

d) V=385mL=0.385L

T=330K

P=1atm

PV=nRT

"n=\\dfrac{PV}{RT} = \\dfrac{1atmx0.385L}{0.08206Latm\/molKx330K}=0.0142mol"

Moles "=\\dfrac{mass}{MM}"

MM "=\\dfrac{mass}{moles}= \\dfrac{1g}{0.0142mol} = 70.34g\/mol"

d)P=4.0x10-2 kpa

V=15mL=0.015L

T=-23+273 = 250K

n=?

Since P is in kpa, R = 8.314kpaL/Kmol

PV=nRT

"n=\\dfrac{PV}{RT}=\\dfrac{4.0x10^-2kpax0.015L}{8.314kpaL\/Kmolx250K} =" 2.887x10-7mol

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