An aqueous solution containing 1.70 g of cane sugar in 100 ml water begins to freeze at -0.093°C.
The molal depression constant of water is 1.86 kg mol-1. Calculate the molecular weight of cane sugar.
Answers
Answer:
the difference between the freezing point of the pure solvent and the freezing point of the solution is given by the freezing-point depression, which can be calculated using the equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
Δ
T
f
=
i
⋅
K
f
⋅
b
a
a
∣
∣
−−−−−−−−−−−−−−−−−−
Here
Δ
T
f
- the freezing-point depression;
i
- the van't Hoff factor
K
f
- the cryoscopic constant of the solvent;
b
- the molality of the solution
The problem provides you with the cryoscopic constant for water
K
f
=
1.86
∘
C kg mol
−
1
Now, sodium iodide is soluble in aqueous solution, which means that it dissociates completely to form sodium cations,
Na
+
, and iodide anions,
I
−
NaI
(
a
q
)
→
Na
+
(
a
q
)
+
I
−
(
a
q
)
Notice that every mole of sodium iodide that is dissolved in solution produces two moles of particles of solute, i.e. ions.
This means that the van't Hoff factor, which tells you the ratio that exists between how many moles of solute you're dissolving and the number of moles of particles of solute that are produced in solution, will be equal to
2
i
=
2
→
one mole of solute dissolved, two moles of ions produced
The molality of the solution is defined as the number of moles of solute present in one kilogram of solvent. Your solution contains
0.5
moles of sodium iodide and
500
g
⋅
1 kg
10
3
g
=
0.5 kg
of water, your solvent. This means that the molaity of the solution will be
b
=
0.5 moles
0.5 kg
=
1 mol kg
−
1
You now have what you need to calculate the freezing-point depression
Δ
T
f
=
2
⋅
1.86
∘
C
kg
⋅
mol
−
1
⋅
1
mol
kg
−
1
Δ
T
f
=
3.72
∘
C
The freezing point of the solution will thus be
Δ
T
f sol
=
0
∘
C
−
3.72
∘
C
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
−
3.7
∘
C
a
a
∣
∣
−−−−−−−−−−−
I'll leave the answer rounded to two sig figs, but keep in mind that your values only justify one sig fig for freezing point of the solution. the difference between the freezing point of the pure solvent and the freezing point of the solution is given by the freezing-point depression, which can be calculated using the equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
Δ
T
f
=
i
⋅
K
f
⋅
b
a
a
∣
∣
−−−−−−−−−−−−−−−−−−
Here
Δ
T
f
- the freezing-point depression;
i
- the van't Hoff factor
K
f
- the cryoscopic constant of the solvent;
b
- the molality of the solution
The problem provides you with the cryoscopic constant for water
K
f
=
1.86
∘
C kg mol
−
1
Now, sodium iodide is soluble in aqueous solution, which means that it dissociates completely to form sodium cations,
Na
+
, and iodide anions,
I
−
NaI
(
a
q
)
→
Na
+
(
a
q
)
+
I
−
(
a
q
)
Notice that every mole of sodium iodide that is dissolved in solution produces two moles of particles of solute, i.e. ions.
This means that the van't Hoff factor, which tells you the ratio that exists between how many moles of solute you're dissolving and the number of moles of particles of solute that are produced in solution, will be equal to
2
i
=
2
→
one mole of solute dissolved, two moles of ions produced
The molality of the solution is defined as the number of moles of solute present in one kilogram of solvent. Your solution contains
0.5
moles of sodium iodide and
500
g
⋅
1 kg
10
3
g
=
0.5 kg
of water, your solvent. This means that the molaity of the solution will be
b
=
0.5 moles
0.5 kg
=
1 mol kg
−
1
You now have what you need to calculate the freezing-point depression
Δ
T
f
=
2
⋅
1.86
∘
C
kg
⋅
mol
−
1
⋅
1
mol
kg
−
1
Δ
T
f
=
3.72
∘
C
The freezing point of the solution will thus be
Δ
T
f sol
=
0
∘
C
−
3.72
∘
C
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
−
3.7
∘
C
a
a
∣
∣
−−−−−−−−−−−
I'll leave the answer rounded to two sig figs, but keep in mind that your values only justify one sig fig for freezing point of the solution.
Explanation:
Answer:
hellooooo
Explanation:
same as above ans