contestada

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 __________.

Respuesta :

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:

[tex]Numer of moles= \frac{Given mass}{Molar mass}[/tex]


[tex]4.0 g of He =\frac{4 g}{4 \frac{g}{mol}}=1 mol[/tex]


[tex]6.5 g of Ar =\frac{6.5 g}{39.95 \frac{g}{mol}} = 0.1627 mol[/tex]


[tex]10.0 g of Ne =\frac{10 g}{20.18 \frac{g}{mol}} = 0.4955 mol[/tex]


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


[tex]Mole fraction of He in substance =\frac{Number of moles of He }{Total number of moles}[/tex]

                                                = [tex]\frac{1}{1.695}=0.6[/tex]