Answer to Question #294684 in General Chemistry for abby

Question

A flask contains a mixture of two gases: NO₂ and Ne. If the flask contains 0.768 mol of NO₂ and 14.8 g of Ne at a total pressure of 1.17 bar, what is the partial pressure of Ne in the flask (in bar?)

TIP: You don't need the volume of the flask nor the temperature to solve this problem.

Answer 1

n = m / Mr

n(Ne) = m(Ne) / Mr(Ne) = 14.8 g / 20.2 g/mol = 0.732 mol

n(total) = n(Ne) + n(NO₂) = 0.732 mol + 0.768 mol = 1.5 mol

$\chi$(Ne) = n(Ne) / n(total) = 0.732 mol / 1.5 mol = 0.488

P_A = $\chi$_A P_total

P_Ne = $\chi$_Ne P_total = 0.488 x 1.17bar = 0.57 bar