Posted by Rose on Thursday, February 26, 2009 at 5:37pm.
I have no idea how to do this problem, but about the mass of CHCl3, don't you just add the atomic mass up from each element to find its total mass?
you don't need mass directly, you are given 100ppb
molality= 100E-6 grams/molmassCHCl3*1kg
remember you need the top mass in grams, not kg (1000E-9ppb*1kg)
noticed the denominator ignored the solute, it is so dilute. Now in molarity, ignore the solute in the volume, and assume denstity is 1.00
molarity=100E-6/molmass*1liter
and so on.
Chopsticks, you are right, you have no idea on this.
100 ppb means 100 parts per billion. That's the same type problem as percent by mass but we are dealing with billions instead of hundreds (percent is hundreds).
So 100 ppb is 100 grams/1,000,000,000 g solution or 100 grams/10^9 grams solution which also means 100 grams/(100 g CHCl3 + 999,999,900 g solvent). You can see that the 100 is so small in comparison to the billion, that we can just ignore that difference for this part of the problem and say we have 100 grams/10^9 g solvent. So 100 g is how many moles? 100/molar mass CHCl3.
Then moles/liter (and the density will be so close to 1.00 g/mL that you can convert grams solvent to mL solvent. After you have molarity, I think the rest will follow for you.
100/119.4 = 0.8375 moles CHCl3/10^9 mL or
0.8375 x 10^-9 moles/mL or
0.8375 x 10^-9 moles/mL x(1000 mL/L) = 0.8375 x 10^-6 moles/L =
8.375 x 10^-7 moles/L which rounds to 8.38 x 10^-7 Molar CHCl3 to 2 significant figures (the 100 has two s.f.). I would call that a very dilute solution. Check my work.
13.4 Expressing Solute Concentration
• Concentration – the ratio of the quantity of
solute to the quantity of solution (or solvent)
��Molarity (M) – the number of moles of solute
per 1 liter of solution
M = (mol of solute)/(liters of solution)
– M is affected by temperature (the solution volume
changes with T, so M changes too)
– The solution volume is not a sum of the solvent and
solute volumes (it must be measured after mixing)
��Molality (m) – the number of moles of solute per
1 kilogram of solvent
m = (mol of solute)/(kilograms of solvent)
– m is not affected by temperature (the amounts of
solute and solvent don’t change with T)
– The solution volume is not needed; m can be
calculated from the masses of solute and solvent
• M and m are nearly the same for dilute aqueous
solutions since 1 L of water is about 1 kg, so (liters
of solution) ≈ (kg of solvent)
Example: Calculate M and m for a solution
prepared by dissolving 2.2 g of NaOH in 55 g of
water if the density of the solution is 1.1 g/mL.
m (molal )
mol solute
1.0 m
kg
1.0mol
0.055 kg water
0.055 mol NaOH
0.055 mol
40 g NaOH
2.2 g NaOH 1 mol NaOH
= = →
= × =
M (molar )
density
Volume mass
1.1 M
L
1.1mol
0.052 L solution
0.055 mol NaOH
52 mL
1.1 g/mL
2.2 g 55 g
= = →
=
+
= =
• Parts of solute by parts of solution
��Parts by mass
– Mass % – grams of solute per 100 grams of
solution → % (w/w)
100%
mass of solution
Mass%= mass of solute ×
100%
volume of solution
Volume%= volume of solute ×
– ppm or ppb – grams of solute per 1 million or 1
billion grams of solution (for trace components)
��Parts by volume
– Volume % – volume of solute per 100 volumes
of solution → % (v/v)– ppmv or ppbv – volume of solute per 1 million
or 1 billion volumes of solution (used for trace
gases in air)
��Mole fraction (X) – ratio of the # mol of
solute to the total # mol (solute + solvent)
mol of solute mol of solvent
mol of solute
+
X =
Example: Calculate the X of NaOH in a solution
containing 2.2 g of NaOH in 55 g of water.
0.018
0.055 mol 3.1 mol
0.055 mol
3.1 mol H O
18 g
0.055 mol NaOH 55 g 1 mol
40 g
2.2 g 1 mol 2
=
+
=
= =
X
• Converting units of concentration
Example: A sample of water is 1.1×10-6 M in
chloroform (CH3Cl). Express the concentration
of chloroform in ppb. (Assume density of 1.0
g/mL)
1.1×10-6 M → 1.1×10-6 mol CH3Cl per 1 L solution
10 ppb 56 ppb
1.0 10 g
5.6 10 g CH Cl
1.0 10 g
mL
1 L 1000 mL 1.0 g
5.6 10 g CH Cl
1 mol CH Cl
1.1 10 mol CH Cl 50.5 g CH Cl
9
3
3
5
3
3
5
3
3
3
-6
× =
×
×
→ × = ×
× × = ×
−
−
solution
solution
Example: What is the molality of a solution of
methanol in water, if the mole fraction of
methanol in it is 0.250?
Assume 1 mol of solution:
→ nmeth = 1 mol × 0.250 = 0.250 mol
→ nwater = 1 - 0.250 = 0.750 mol
18.5 m
0.0135 kg H O
0.250 mol methanol
0.0135 kg H O
10 g
1 kg
1 mol H O
0.750 mol H O 18.0 g H O
2
3 2
2
2
2
= =
× × =
m
Example: What is the molality of a 1.83 M
NaCl solution with density of 1.070 g/mL?
Assume 1 L (103 mL) of solution:
→ nNaCl = 1.83 mol
1.90 m
0.963 kg H O
1.83 mol NaCl
1070 g 107 g 963 g 0.963 kg
107 g
1 mol
a 1.83 mol 58.44 g NaCl
1070 g
1 mL
10 mL 1.070 g
2
3
= =
= − = =
= × =
= × =
m
mass of water
mass of N Cl
mass of solutionExample: What is the molarity of a 1.20 m KOH
solution in water having density of 1.05 g/mL?
Assume 1 kg (1000 g) of solvent (H2O):
→ nKOH = 1.20 mol
1.18 M
1.02 L
1.20 mol KOH
1016 mL 1.02 L
1.05 g
1067 g 1 mL
1000 g 67.3 g 1067 g
67.3 g
1 mol
1.20 mol 56.1 g KOH
= =
= × = =
= + =
= × =
solution
M
volume of solution
mass of solution
mass of KOH