A solution of hexane in the solvent (CH3)2CO has a mole fraction of CH3(CH2)4CH3 of 0.114. What mass (g) of the solution must be taken if 1690 g of hexane is needed?

Question

A solution of hexane in the solvent (CH3)2CO has a mole fraction of CH3(CH2)4CH3 of 0.114. What mass (g) of the solution must be taken if 1690 g of hexane is needed? 
Molar Mass (g/mol) 
CH3(CH2)4CH3 100.21 
(CH3)2CO 58.05 
 
Density (kg/L):  
CH3(CH2)4CH3 0.6838 
(CH3)2CO 0.7899 
 
Name/Formula: 
hexane 
CH3(CH2)4CH3 
acetone 
(CH3)2CO

DrBob222 · Answer

0.114 = X(hexane)/X(hexane)+X(acetone)
X(hexane) = 1690 g hexane/100.21
X(acetone) = grams acetone/58.05
You have only one unknown. Solve for grams acetone. Then g acetone + 1690 = total grams.Check my thinking. Check my work. Post your work if you need further assistance.