`O_(2)` and `SO_(2)` gases are filled in ratio of 1 : 3 by mass in a closed container of 3 L at temperature of `27^(@)C`. The partial pressure of `O_(2)` is 0.60 atm, the concentration of `SO_(2)` would be
A
0.36
B
0.036
C
3.6
D
36
Text Solution
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The correct Answer is:
b
(b) We have `P_(O_(2))=X_(O_(2))xxP_("total")` `:." " 0.60=(n_(O_(2)))/(n_(O_(2))+n_(SO_(2)))xxP_("total")` Let no. of moles of `O_(2)` are `x,` `0.60=(x//32)/(x//32+3x//64)xxP_("total")` `:. 0.60=0.4xxP_("total")` `:. " "P_("total")=(0.60)/(0.4)=1.5"atm"` `:'P_(O_(2))+P_(SO_(2))=1.5"atm"` `implies P_(SO_(2)) =1.5-P_(O_(2))=1.5-0.60=0.9"atm"` `:. "Concentration of "SO_(2)=(0.9)/(0.0821xx300)` `=0.036`
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