## Chemistry and Chemical Reactivity (9th Edition)

Per kg of solvent: The number of moles of solute $10.7\ mol/kg\times 1\ kg=10.7\ mol$ The number of moles of solvent: $1000\ g\div 18.015\ g/mol=55.51\ mol$ a) Mole fraction: $10.7/(10.7+55.51)=0.162$ Mass of solute: $10.7\ mol\times 40.00\ g/mol=428\ g$ b) Weight percent: $428/(1000+428)\times 100\%=30\%$ The volume of the solution: $(1000+428)g\div1.33\ g/cm^3\times 1\ L/1000\ cm^3=1.074\ L$ c) Molarity: $10.7\ mol\div 1.074\ L=10.0\ M$