3.2 g of oxygen (At. weight = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12 litre flask at 0°C. The total pressure of the gas mixture will be

(a) 1 atm

(b) 4 atm

(c) 3 atm

(d) 2 atm

choose and explain

Question

3.2 g of oxygen (At. weight = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12 litre flask at 0°C. The total pressure of the gas mixture will be

(a) 1 atm

(b) 4 atm

(c) 3 atm

(d) 2 atm

choose and explain

KUMAR PRASSU · Answer

There is a total of 0.2 moles of gas =>n=0.2 volume of container is 1.12L=>V=1.12L temp. is 273 K=>T=273K there fore by PV=nRT pressure=4atm there fore option 2 is correct hope u like this PLS APPROVE

Vikas TU · Answer

You are dealing with diatomic oxygen ( molar mass 32g/mol ) and diatomic hydrogen ( 2g/mol ) Mol O2 in flask = 3.2/32 = 0.1 mol Mol H2 in flask = 0.2/2 = 0.1 mol H2 Total moles = 0.2mol Calculate pressure using gas equation: PV = nRT P * 1.12 = 0.2*0.082057*273 P = 4.480/1.12 P = 4.000 atm Pressure = 4atm - choice b) is correct. Plz Approve!

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3.2 g of oxygen (At. weight = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12 litre flask at 0°C. The total pressure of the gas mixture will be (a) 1 atm (b) 4 atm (c) 3 atm (d) 2 atm choose and explain

3.2 g of oxygen (At. weight = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12 litre flask at 0°C. The total pressure of the gas mixture will be

(a) 1 atm

(b) 4 atm

(c) 3 atm

(d) 2 atm

choose and explain

2 Answers

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