Question

The mole fraction of he in a gaseous solution prepared from 4.0 g of he, 6.5 g of ar, and 10.0 g of ne is __________.

Answer 1

Given that there is 4.0 g of He, 6.5 g of Ar, and 10.0 g of Ne :

So, first all these masses are converted into moles:

$Numer of moles= \frac{Given mass}{Molar mass}$

$4.0 g of He =\frac{4 g}{4 \frac{g}{mol}}=1 mol$

$6.5 g of Ar =\frac{6.5 g}{39.95 \frac{g}{mol}} = 0.1627 mol$

$10.0 g of Ne =\frac{10 g}{20.18 \frac{g}{mol}} = 0.4955 mol$

Total Number of moles  = 1 + 0.1627 + 0.4955 = 1.658 mol

$Mole fraction of He in substance =\frac{Number of moles of He }{Total number of moles}$

                                                = $\frac{1}{1.695}=0.6$