Question

2 g of He, 3 g of N; and 4 g of Ar were introduced into a 15 dm

Answer 1

Moles of He $=\frac{2}{4.00}=0.5moles$

Moles of N$=\frac{3}{14.00}=0.214$

Moles Ar$=\frac{4}{39.948}=0.102moles$

Average moles $=\frac{0.5+0214+0.102}{3}=0.272moles$

Molarity $=\frac{0.272}{15}=0.018M$