Question

A gaseous mixture contains three gases A, B and C with a total number of 10 moles and total pressure of 10 atm. The partial pressure of A, B and C are 6 atm, 2 atm and 2 atm respectively and if C has molecular weight of 20 g/mol, then what will be the weight of C present in the mixture?​

Answer 1

Answer:

C= 10 - (6+2)

= 2

Therefore C = 20g

= 20 x 2

= 40g Ans

Explanation:

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