fraction of NaOH is 0.159.

A 10.5 mol/kg solution of NaOH contains 10.5 mol NaOH and 1 kg of water.

##”Moles of water” = 1000 color(red)(cancel(color(black)(“g H”_2″O”))) × (“1 mol H”_2″O”)/(18.02 color(red)(cancel(color(black)(“g H”_2″O”)))) = “55.49 mol H”_2″O”##

The formula for mole fraction ##x## is

##color(blue)(|bar(ul(color(white)(a/a)x = “moles of component”/”total moles”color(white)(a/a)|)))” “##

##”Total moles” = “moles of NaOH” + “moles of water” = “10.5 mol + 55.49 mol” = “65.99 mol”##

∴ ##x_”NaOH” = (10.5 color(red)(cancel(color(black)(“mol”))))/(65.99 color(red)(cancel(color(black)(“mol”)))) = 0.159##