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For 10 Minutes each at 27 Deg C from two identical holes nitrogen and an unknown gas are leaded in to a common vassel of 3 L Capicity. The resulting pressure is 4.18 Bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknowen gas.

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3 years ago

```							Here, V = 3L , T = 300 K , R = 0.0821 Lit atmJ/K mol, P = 4.18 BarSo, total moles of solution n =PV/RT  = 4.18 x 3 /(0.0821 x 300)  = 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
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3 years ago
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