# What is the volume and L of a 9.21 g sample of oxygen gas at 40.7°C that exerts 886.7 mmHg of pressure? ?

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- jacob sLv 74 weeks ago
1) 760 mm Hg = 1 atm

886.7 mm Hg = 886.7/760 atm= 1.17 atm

V = ?

P = 1.17 atm

n = 9.21g O2 x (1 mole O2/32 g O2)= .2878

R = 0.0821 L. atm/mol/K

Formula: ideal gas equation, PV=nRT

Temperature in oC = 40.7°C + 273= 313.7°K

(1.17atm)V=.2878( 0.0821 L. atm/mol/K)( 313.7°K)

V =.2878( 0.0821 L. atm/mol/K)( 313.7°K)/(1.17atm)

V= 6.33L

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