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
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
A solid consists of a mixture of NaNO3 and Mg(NO3)2. When 6.50 g of the solid...
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