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Grade: 10
        
I want the answer of the question . Please it is urgent 
one year ago

Answers : (1)

Arun
22992 Points
							
Here, V = 4L , T = 273 K , R = 0.0821 Lit atmJ/K mol, P = 2.8 BarSo, total moles of solution n =PV/RT = 2.8 x 4 /(0.0821 x 273) = 0.5 molesAs moles og nitrogen gas = 0.4 Hence, no. of moles of unknown gas = 0.5 -0.4 = 0.1 molesNow, from Grahm`s law of diffusion,Rate of effusion of nitrogen ( rN) = no. of moles/ time taken = 0.4/ 10 = 0.04 mol/minRate of effusion of unknown gas (runknown ) =0.1/10 = 0.01 mol/minFurther, r n/r unknown=sqrt m unknown/ m n= 0.04/0.01 = 4, where m = molar masses Or, squaring both side we get, (munknown ) / (mN ) =16Or, (munknown ) = (mN ) x 16 Since, molar mass of nitrogen gas (N2 ) = 28So, (munknown ) = 28x16 = 448 gm/mol
one year ago
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