What mass of carbon monoxide in grams is present in a room measuring 9.5 x 12.5 x 19.8ft ?

Question

The concentration of carbon monoxide in an urban apartment is 48 ug/m^3.

Answer 1

First calculate the volume of the room as 9.5*12.5*19.8= 2351.25 ft^3



then you need to convert the feet measurement into metres



There are 3.2808 ft in one metre, and one cubic metre is equal to 3.2808*3.2808*3.2808= 35.315 cubic feet





the room volume is therefore 2351.25/35.315=66.58 m^3





The amount of carbon monoxide in room is 48*66.58= 3195.84 micrograms



But 1 microgramme is equivalent to 0.000001



The amount of carbon dioxide is 3195.84*0.000001= 0.00319584 grams

Answer 2

formulation for concentration is: concentration=Mass/quantity the quantity of the room is 39,seventy two m3. that's more suitable with the concentration of CO interior the room (40 six.7 mg/m3) and from that the mass of CO is a million.854gr.

Answer 3

let me try in another solution

using mark room size of 66.58m3

density of air at 20C is 1.2041 kg/m3

mass of air = (66.58)(1.2041) = 80.17kg



converting %v/v to %w/w

%v/v of CO in air = 1E-6 %

mw of CO in air = (%v/v CO)(28.01) = (1E-6)(28.01)/100 = 2.80E-7

%w/w of CO = (mw of CO in air)(100) / (mw of air)

%w/w of CO = (2.80E-7)(100) / (28.972)

%w/w of CO = 9.668E-7 %



mass of CO = (9.668E-9)(80.17) = 7.751E-7 kg or 7.75E-4 grams

my answer is astronomically wrong, sorry



if using your figure 48E-6 g/m3

mass of CO = (66.58)(48E-6) = 0.003 grams