Question

A gaseous mixture of three gases A. B and C has a pressure of 10 atm. The total number of moles of all the gases is 10. If the partial pressures of Aand B are 3.0 and 1.0 atm respectively and if C has a mol/wt. of 2.0. What is the weight of C in g present in the mixture ?

• A. 6
• B. 3
• C. 12
• D. 8

Answer 1

Answer: (C)

12

Answer 2

Total pressure of mixture of gases A, B and C = 10 atm. Number of total moles = 10 Partial pressure of A = $\frac{No. of \, moles \, of \, A}{10} \times 10$ $\therefore$ 3 = No. of moles of A $\because$ Partial pressure of B = $\frac{No.of \, moles \, of \, B}{10} \times 10$ = No. of moles of B. $\therefore$ No. of moles of C = 10 - (3 +1) = 6. Now 1 mole of C = 2g $\therefore$ 6 mol of C = 2 x 6 = 12 g