The weight of gaseous mixture containing 6.02×10^23molecules of nitrogen and 3.01×10^23molecuels of sulphur dioxide are

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Solution

We know that 6.02×10^23 = 1mol
slly 3.01×10^23 = .5 mol
therefore, weight of 1 mol of N2 = 2 x 14= 28g
weight of .5 mol of SO2 = molecular weight of SO2 × number of moles
= 64 × .5 = 32 g

therefore the total weight = 28 + 32 g = 60 g

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A flask contains 3.2 g of sulphur dioxide. Calculate the followings:

i) The number of molecules of sulphur dioxide at STP.

ii) The volume occupied by 3.2 g of sulphur dioxide at STP.

(S = 32, O = 16)

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