Question 1:
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)
Solution:
Weight of 200 ml of HNO3 is 200 (volume) x 1.48 (density) = 296 g
Weight of 200 ml of H2SO4 is 200 (volume) x 1.9 (density) = 380 g
So, the weight of the solution is 296 + 380 = 676 g
Question 2:
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.
Solution:
Mole fraction of ethanol:
x = molesethanol/ (molesethanol + moleswater) = 0.05
Taking 100 moles of solution as the basis the solution is formed by 5 moles of ethanol and 95 moles of water.
As 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
Question 3:
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.
Solution:
The volume of the resulting solution is 1000 + 750 = 1750 ml.
In 1 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
So, the final solution contain 18 + 24.75 = 42.75 g NaCl
Hence, 42.75 : 1750 = x : 100
x = 2.44%
Question 4:
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 g/ml. Calculate the volume of the resulting solution.
Solution:
Formula: Density = Mass / volume
Mass = Density X Volume
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.
Question 5:
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.
Solution:
Calculate the volume occupied by 100 grams of solution:
Density = Mass/volume
Volume = Mass/density = 100/1.834 = 54.52 ml
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
Question 6:
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.
Solution:
From density we can calculate the volume occupied by 100 g of solution as follows,
Density = mass / volume
V= 100/1.11 = 90.09
So, the solution contains 20 gr of the acid in 90.09 ml,
The volume containing 10 g of the acid is 90.09/2= 45.045.
Question 7:
When 300 ml of water is 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.
Solution:
The resulting volume:
400 - (4)(400)/100= 400 - 16 = 384 ml
The grams of the H2SO4 solution:
(volume)(d) = (100) (1,2) = 120 grams
The new value of density:
(120 + 300)/384 = 1.09
Question 8:
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).
Solution:
Calculate the weight of the solution of NaOH using the below formula,
d = mass/volume
Mass = (volume) (d) = (150) (1.70) = 255 grams
The final weight after adding water is,
255 + 350 grams water = 605 grams
The final volume, assuming no variation on mixing, will be 500 ml and so,
d = 605/500 = 1.211 gr/ml
Question 9:
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.
Solution:
According to the reaction, for 5 moles of oxalate are required 2 moles of permanganate,
Molesoxalate/Molespermanganate = 5/2 = 2.5
Here, the moles of oxalate are 0.01 so, 0.01/Molespermanganate = 2.5
from which Molespermanganate = 0.01/2.5 = 0.004
As the solution of permanganate is 0.1 M,
0.1 moles : 1000 = 0.004 : x
x = 0.004/0.0001 = 4O ml
Question 10:
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.
Solution:
By indicating with V1 and V2 the volumes of the two solutions , it must be:
Here,
g1 + g2 = 0.05 (V1) + 0.2 (V2) = 30 grams.
By substituting V2 with 300 - V1 we have that
So,
V1 = 30/0.15 = 200 ml
V2 = 300 - 200 = 100 ml
Question 11:
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.
Solution:
By indicating with V1 and V2 the volumes (in liters) of the solution 12M and 3M respectively, it must be:
V1 + V2 = 1 liter
The final solution must result 6 M then,
Moles 1 + Moles 2 = 6
MiVi = MfVf we have that
Moles1 = (V1)(12)
Moles2 = (V2)(3)
12.V1 + 3.V2 = 6
By substituting V1 with 1-V2 we have
12 (1-V2) + 3V2= 6
12 - 12V2 + 3V2= 6
12-6 = 12V2 - 3V2 = 9 V2
From which.,
V2 = 6/9 = 0.666 liters = 666 ml
V1 = 1000 - 666 = 334 ml
Question 12:
Calculate the molarity of a solution containing 100 gr of NaCl (Mw = 58.5) in 1.5 liters of water.
Solution:
Formula : Moles = grams of solute /mw = 100/58.5 = 1.70
1.70 : 1500 = M : 1000
M = (1.70 ) (1000)/1500 = 1.14
Question 13:
You need to withdraw 200 grams of a KCl solution having density = 1.16 g/ml. How many ml of the solution will you withdraw ?
Solution:
Formula : Density = mass/volume
Volume = mass/d
= 200/1.16 = 172,4 ml
Question 14:
Calculate the concentration expressed in grams% (w/v) of a 1 M solution of HCl (Mw=36).
Solution:
A solution 1 M contains 1 mole of solute /liter,
So, 36 grams / liter of HCl, how many grams in 100 mL?
36 gr : 1000 ml = x : 100 ml
x = (36)(100)/(1000) = 3,6 g (3.6%)
Question 15:
How many grams of Na2SO4 (Mw = 142) are needed to prepare 5 liters of a solution 0.1 M ?
Solution:
0.1 M contains 0.1 moles/liter
One mole of Na2SO4 = 142 grams,
So, the solution contains 142/10 = 14.2 g of salt for a liter of solution.
In 5 liters: (14.2) (5) = 71 grams
Question 16:
Calculate the ml of a 1 M solution of NaCl needed to prepare 100 ml of a 0.2 M solution.
Solution:
Formula : MiVi = MfVf
Substitute the know and given values,
(x) (1) = (100)(0.2)
x = 20 ml
Question 17:
Calculate the grams of HCl (mw = 36) present in two liter of a solution 0.3 M.
Solution:
Molarity = Moles / Volume
A solution 0.3 M contains 0.3 moles/liters so two liters will contain (0.3) (2) = 0.6 mole.
Number of moles = mass of solute / molar mass
Mass of solute = (moles) (molar mass) = (0.6)(36) = 21.6 grams
Question 18:
2 liter of a solution contain 1 mole of HCl. Calculate the molarity of the solution.
Solution:
Molarity = Moles / Volume
M = 1 mole / 2 liters
M = 0.5
Question 19:
Calculate the grams of NaCl (mw = 58) contained in 30 ml of a 0.2 M solution.
Solution:
0.2 M solution contains 0.2 moles/liter
Moles = grams of solute / 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 : 30
x = (11.6) (30)/1000 = 0.348 grams of NaCl
Question 20:
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.
Solution:
Moles present in 20 gr of NaOH :
Moles = mass/mw = 20/40 = 0.5 moles
As Molarity is the number of moles of solute in 1 liter of solution, the solution is 0.5 molar.
Question 21:
500 ml of a solution contains 20 grams of NaOH (mw = 40). Calculate the molarity of the solution.
Solution:
20 g of NaOH is 0.5 moles which are contained in 0.5 liter of solution.
In one liter of solution there will be (2) X (0.5) = 1 mole and hence the solution is 1 M.
Question 22:
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.
Solution:
i) Moles = grams/molecular weight
(moles) (mw) = (36) (0.2) = 7.2 grams
ii) A solution 1 M contains 1 mole/liter thus 36 grams/liters.
It follows that for 500 ml we need:
36/2 = 18 grams
Question 23:
How many water you have to add to 450 ml of a solution 0.3 M to obtain a concentration 0.25 M ?
Solution:
Formula: MiVi = MfVf and thus
(0.45)(0.3) = (0.25)(Vf)
Vf = 0.54 liter = 540 ml
So, the water to add is 540 - 470 = 70 ml.
Question 24:
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 ?
Solution:
To prepare 2 liters of a 1 M solution we need 2 moles of NaOH, i.e. 80 grams.
So, we need to calculate the volume of the 0.2% solution of NaOH containing 80 grams of NaOH.
Formula: M1V1 = M2V2
0.2 : 100 = 80 : v
v = (80)(100)/0.2 = 40000 ml = 40 liters
Therefore it is not possible to prepare the solution.
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