# Question 5373c

May 21, 2015

The final volume of the solution is of no importance for determining the mole fraction of sodium iodide in the solution, so you can ignore it altogether.

You can think of the mole fraction of sodium iodide in the solution as the ratio between the moles of sodium iodide and the total number of moles the solution contains, i.e. moles of solute and moles of solvent.

Use sodium iodide and water's molar masses to determine how many moles of each you have

76.3cancel("g") * "1 mole NaI"/(149.9cancel("g")) = "0.509 moles" $N a I$

and

545cancel("g") * "1 mole water"/(18.02cancel("g")) = "30.24 moles" ${H}_{2} O$

This means that the total number of moles present in the solution will be

${n}_{\text{total}} = {n}_{N a I} + {n}_{{H}_{2} O}$

${n}_{\text{total" = 0.509 + 30.24 = "30.75 moles}}$

Therefore, the mole fraction of sodium iodide will be

${\chi}_{N a I} = {n}_{N a I} / {n}_{\text{total}}$

chi_(NaI) = (0.509cancel("moles"))/(30.75cancel("moles")) = color(green)("0.0166")#