### Solutions | Concentration terms | Worksheet

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:

 V1 + V2 = 300 ml The final solution (10%) must contain 30 gr of NaCl thus g1 + g2 = 30 grams

Here,

 g1 = (5) (V1)/100 = 0.05 (V1)  and g2 = (20) (V2)/100 = 0.2 (V2)

g1 + g2 = 0.05 (V1) + 0.2 (V2) = 30 grams.

By substituting V2 with 300 - V1 we have that

 0.05 (V1) + 0.2 (300 - V1) = 30 ; 0.05 V1 + 60 - 0.2V1 = 30 ; 0.15V1 = 60 - 30

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 = MfV

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. 