Finding Molarity of a Solution That's in Grams
Concentration Units: Solved Problems
1. Is it possible to obtain 2 liters of a solution of NaOH (Mw = 40) 1 M by diluting a solution containing 0,2 grams of NaOH in 100 ml of solution ?
In order to prepare 2 liters of a 1 M solution we need 2 moles of NaOH, i.e. 80 grams.
Therefore, we need to calculate the volume of the 0.2% solution of NaOH containing 80 grams of NaOH.
This volume can be calculated as follows:
0,2 : 100 = 80 : vv = (80)(100)/0.2 = 40000 ml = 40 liters
Therefore it is not possible to prepare the solution.
2. How many water you have to add to 450 ml of a solution 0.3 M to obtain a concentration 0.25 M ?
This problems can be easily solved by rembering that MiVi = MfVf and thus
(0.45)(0.3) = (0.25)(Vf)
(0.45)(0.3)
Vf = ------------- = 0.54 liter = 540 ml
Therefore the water to add is 540 - 470 = 70 ml. Alternatively we can observe that the initial concentration is 0.3/0.25 = 1.2 times more concentrated than the final one. It follows that the volume of the final (diluted) solution should be (450) (1.2) = 540 ml.
3. The molecular weight (mw) of HCl is 36, calculate: i) the grams of HCl contained in 0.2 moles; ii) the grams of HCl needed to prepare 500 ml of a solution 1 M.
i) moles are given by grams/molecular weight, thus
(moles) (mw) = (36) (0.2) = 7.2 grams
ii) a solution 1 M contain 1 mole/liter thus 36 grams/liters. It follows that for 500 ml we need:
36/2 = 18 grams
4. 500 ml of a solution contains 20 grams of NaOH (mw = 40). Calculate the molarity of the solution.
20 gr of NaOH correspond to 0.5 moles which are contained in 0.5 liter of solution. In one liter of solution there will be (2)(0.5) = 1 mole and thus the solution is 1 M.
5. A solution of NaOH (mw = 40) is prepared by dissolving 20 grams of the base in enough water to make one liter of solution. Calculate the molarity of this solution.
the moles present in 20 gr of NaOH are:
moles = gr/mw = 20/40 = 0.5 moles
Since molarity is the number of moles in 1 liter of solution, the solution is 0.5 molar.
6. Calculate the grams of NaCl (mw = 58) contained in 30 ml of a 0.2 M solution.
A 0.2 M solution contains 0.2 moles/liter and remembering that moles = grams/mw
grams = (moles) (Mw) = (0.2) (58) = 11.6 grams of NaCl
Now we can compute the grams in 30 ml with the following proportion
11.6 : 1000 = x : 30x = (11.6) (30)/1000 = 0.348 grams of NaCl
7. 2 liter of a solution contain 1 mole of HCl. Calculate the molarity of the solution.
M = 1 mole / 2 liters = 0.5
8. Calculate the grams of HCl (mw = 36) present in two liter of a solution 0.3 M.
A solution 0.3 M contain 0.3 moles/liters thus two liters will contain (0.3) (2) = 0.6 mole. Since number of moles = gr/mw it follows that gr = (moles) (mw) = (0.6)(36) = 21.6 grams.
9. Calculate the ml of a 1 M solution of NaCl needed to prepare 100 ml of a 0.2 M solution.
By rembering that MiVi = MfVf we have that
(x) (1) = (100)(0.2); x = 20 ml
10. How many grams of Na2SO4 (Mw = 142) are needed to prepare 5 liters of a solution 0.1 M ?
A solution 0.1 M contains 0.1 moles/liter, since one mole of Na2SO4 corresponds to 142 grams then the solution contains 142/10 = 14.2 grams of the salt for liter of solution. For 5 liters:
(14.2) (5) = 71 grams
11. Calculate the concentration expressed in grams% (w/v) of a 1 M solution of HCl (Mw=36).
A solution 1 M contains 1 mole/liter, i.e. 36 grams/liter of HCl thus
36 gr : 1000 ml = p : 100 mlp = (36)(100)/(1000) = 3,6 g (3.6%)
12. You need to withdraw 200 grams of a KCl solution having density = 1.16 g/ml. How many ml of the solution you will withdraw ?
d = mass/volumeVolume = mass/d = 200/1.16 = 172,4 ml
13. Calculate the molarity of a solution containing 100 gr of NaCl (Mw = 58.5) in 1.5 liters of water.
N° moles = grams/mw = 100/58.5 = 1.70
1,70 : 1500 = M : 1000
M = (1.70 ) (1000)/1500 = 1,14
14. One liter of a solution 6 M HCl is to be prepared by mixing two different solutions of the acid:
1. HCl 12 M
2. HCl 3 M
Assuming that there is no variation of the volume during the mixing, calculate the ml of solution 1 and solution 2 to be mixed.
By indicating with V1 and V2 the volumes (in liters) of the solution 12M and 3M respectively, it must be:
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Rembering that MiVi = MfVf we have that
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By substituting V1 with 1-V2 we have
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From which
V2 = 6/9 = 0.666 liters = 666 mlV1 = 1000 - 666 = 334 ml
15. The preparation of 300 ml of a 10% (w/v) solution of NaCl is to be accomplished by mixing two different solutions of the salt:
1. NaCl 5%
2. NaCl 20%
Assuming that there is no variation of the volume during the mixing, calculate the ml of 1 and 2 needed.
By indicating with V1 and V2 the volumes of the two solutions , it must be:
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where
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Thus gr1 + gr2 = 0.05 (V1) + 0.2 (V2) = 30 grams. By substituting V2 with 300 - V1 we have that
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From which
V1 = 30/0.15 = 200 mlV2 = 300 - 200 = 100 ml
16. Potassium permanganate (KMnO4) reacts with oxalate ion (C2O4 --) according to the following reaction:
2MnO4 - + 5C2O4 -- + 16H+ = 2Mn++ + 10CO2 + 8H2O
Calculate the milliliters of 0.10 M KMnO4 that react completely with 0.01 moles of oxalate.
According to the reaction, for 5 moles of oxalate are required 2 moles of permanganate, i.e.,:
Molesoxalate/Molespermanganate = 5/2 = 2.5
in our case the moles of oxalate are 0.01 thus
0.01/Molespermanganate = 2.5
from which
Molespermanganate = 0.01/2.5 = 0.004
Since the solution of permanganate is 0.1 M:
0.1 moles : 1000 = 0.004 : xx = 0.004/0.0001 = 4O ml
17. Calculate the density of a solution obtained by mixing 150 ml of NaOH (d = 1.70 gr/ml) and 350 ml of distilled water (d = 1 gr/ml).
As first let's calculate the weight of the solution of NaOH
d = mass/volumemass = (volume) (d) = (150) (1.70) = 255 grams
The final weight after water addition will be:
255 + 350 grams water = 605 grams
the final volume, assuming no variation on mixing, will be 500 ml and then:
d = 605/500 = 1.211 gr/ml
18. When 300 ml of water are added to 100 ml of H2SO4 solution (d = 1.2 g/ml) a contraction of the volume of 4% is observed. Calculate the density of the final solution.
The resulting volume will be:
400 - (4)(400)/100= 400 - 16 = 384 ml
The grams of the H2SO4 solution are:
(volume)(d) = (100) (1,2) = 120 grams
Thus the new value of density will be
(120 + 300)/384 = 1.09
19. A solution of HNO3 is 20% by weight (i.e., 20 g of acid in 100 g of solution) and has a density of 1.11 gr/ml. Calculate which volume of the solution contains 10 g of HNO3.
From the definition of density we can calculate the volume occupied by 100 g of solution:
V= 100/1.11 = 90.09
Thus the solution contains 20 gr of the acid in 90.09 ml, then the volume containing 10 g of the acid is 90.09/2= 45.045.
20. A solution of H2SO4 (density = 1.834 gr/ml) contains 95 g of the acid per 100 grams of solution. Calculate the volume of the solution containing 38 gr of the acid.
As first let's calculate the volume occupied by 100 grams of solution:
density = mass/volume
volume = mass/density
volume = 100/1.834 = 54.52 ml
Now we can calculate the volume containing 38 grams of the acid as follows:
95 grams: 54.52 ml = 38 : V
V = (54.52) (38)/ (95) = 21.8 ml
21. A solution of KOH (density = 1.4 gr/ml) was obtained by mixing 100 ml of a solution of KOH (density = 1.5 gr/ml) with 100 ml of another solution having density = 1.1 gr/ml. Calculate the volume of the resulting solution.
The weight of the KOH solution with density = 1.5 is (100) (1.5) = 150 g
The weight of the KOH solution with density = 1.1 is (100) (1.1) = 110 g
Thus the weight of the final solution is 260 g , from which:
volume = 260 (grams)/1.4 (density) = 185.7 ml.
22. A solution of NaCl was obtained by mixing two different solutions of the salt:
1. one liter of a solution containing 1.8 g/100 ml of solution
2. 750 ml of another solution containing 3.3 g/100 ml of solution.
Calculate the concentration of the resulting solution expressed as grams/100 ml of solution.
The volume of the resulting solution is 1000 + 750 = 1750 ml. In one liter of the solution 1.8 g/100 there are:
1.8 : 100 = x : 1000; x = 18 gr NaCl
In 750 ml of solution containing 3.3 g/100 ml of solution there are
3.3 : 100 = x : 750; x = 24.75 gr NaCl
Thus the final solution contain 18 + 24.75 = 42.75 g NaCl thus
42.75 : 1750 = x : 100
x = 2.44%
23. Calculate the molality and molarity of a solution of ethanol in water, by knowing that the mole fraction of ethanol is 0.05 and the density of the solution is 0.997 grams/ml.
The mole fraction of ethanol is given by:
x = molesethanol/ (molesethanol + moleswater) = 0.05
Taking 100 moles of solution as basis the solution is formed by 5 moles of ethanol and 95 moles of water. Since the molality is the number of moles of solute in 1 kg of solvent we need to calculate the kg of water forming the solution:
95 x 18 = 1710 grams water = 1.710 kg water
m = 5/1.710 = 2.92
Molarity is defined as the number of moles in 1 liter of solution. The solution is formed by 1.710 kg of water and 5 * 44 = 220 grams = 0.22 kg of ethanol,i.e.
weightsolution = 1.710 + .220 = 1.93 kg
from the value of density we can calculate the volume of the solution
volume = 1.93/0.997 = 1.935 and M = 5/1.935 = 2.58
24. Calculate the weight of a solution prepared by mixing:
200 ml of a solution of HNO3 (d = 1.48 g/ml)
200 ml of a solution of H2SO4 (d = 1.9 g/ml)
the weight of 200 ml of HNO3 is 200 (volume) x 1.48 (density) = 296 g
the weight of 200 ml of H2SO4 is 200 (volume) x 1.9 (density) = 380 g
Thus the weight of the solution is
296 + 380 = 676 g
Finding Molarity of a Solution That's in Grams
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