Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes :

Explanation

The relation between Molarity and mole fraction is given by
`(X_(2))=("Molarity Molecular wt(s)")/("Molarity (molecular wr (theta)-Mol wt (s) + 1000 d")`= `(MxxM_(1))/(M(M_(1)-M_(2))+1000 d)`
Given data, `xx_(2)`0.040 M=?
`M_(1)=18(H_(2)O),M_(2)=46(C_(2)H_(5)OH),Cl=1`
0.040 = `(Mx18)/(M(18-46)+1000 xx 1)=(Mxx18)/(-28M+1000)`
18 M = -0.040 xx M xx 28+ 0.040 X 1000
18 M =-1.12M +40
(18+1.12)M=40
19.12 M = 40
M= `(40)/(19:12)=2.09 M:.` Molarity of Ethanol = 2.09 M
Given data
Molar mass of Argon = `overset("Isotope -I mass"_+"Isotope II mass "_+"Isotope -III mass")(xx "Relative abudance X Relative abudance X Relative abudance")/(100)`
`(35.96755xx0.337+37.96272x0.063+39.9624+99.6)/(100)`
`(12.12106+2.39165+3980.25)/(100)=39.948`