Question

A solid consists of a mixture of NaNO3 and Mg(NO3)2. When 6.50 g of the solid is dissolved in 50.0 g of water, the freezing point of the solution is lowered by 5.26°C. What is the composition by mass of the solid?

----- g NaNO3

----- g Mg(NO3)2

Answer 1

A solid consists of a mixture of NaNO3 and Mg(NO3)2. When 6.50 g of the solid is dissolved in 50.0 g of water, the freezing point of the solution is lowered by 5.26°C. What is the composition by mass of the solid?

Sol:

Using the freezing point depression constant:

ΔT= Kf * M (Hire,M = molality)

-5.26 = (-1.86) *M

M = 5.26 / 1.86

M = 2.828 molal solution

find the moles of ions in with 50 grams of water:
50 g H₂O * 2.828 moles of ions / 1000 g of water = 0.1414 moles of ions ........(*)

Assume grams of NaNO₃, let say = X

Given that

A solid consists of a mixture of NaNO3 and Mg(NO3)2. When 6.50 g of the solid is dissolved in 50.0 g of water

grams of Mg(NO3)2 =   (6.50 - X)

knowing that each mole of NaNO₃ releases 2 moles of ions.
& that each mole of Mg(NO₃)₂ releases 3 moles of ions

molar mass NaNO3 =84.9947 g/mol

molar mass Mg(NO₃)₂ =148.3148 g/mol

convert those grams of compounds into moles of ions , using molar masses

For NaNO3

(X g NaNO3) ( 2moles of ions) / (84.99 g/mol NaNO3) = 0.02353 X moles of ions from NaNO3 ......(1)

For Mg(NO3)2

(6.50 - X) g Mg(NO3)2 (3 moles of ions) / (148.32 g/mol Mg(NO3)2) =

(6.50 - X) g Mg(NO3)2 (0.02023) = 0.1315 - 0.02023 X moles from Mg(NO3)2 ----(2)

from (*) and (1 ) and(2)

these moles add up:
(0.02353 X) + (0.1315 - 0.02023 X ) = 0.1414 moles of ions found by freezing point depression

0.02353 X + 0.1315 - 0.02023 X = 0.1414

0.02353 X + 0.1315 - 0.02023 X = 0.1414

0.0033 X = 0.0099

X = 3 gram

X = 3 grams of NaNO3

and

6.50 -X = 3.5 gram of Mg(NO3)2