Task number 4 on the topic "Solutions"

Examples of problem solving

1.1. PERCENTAGE CONCENTRATION

Example 1

a) Determine the mass fraction (%) of potassium chloride in a solution containing 0.053 kg KCl in 0.5 l of a solution whose density is 1063 kg / m 3 .

Decision:

The mass fraction ω or C% shows how many mass units of the solute are contained in 100 mass units of the solution. Mass fraction is a dimensionless quantity, it is expressed in fractions of a unit or percentage:

where ω A – mass fraction (%) of the dissolved substance;

m A is the mass of the dissolved substance, g;

m solution – the mass of the solution, g.

The mass of the solution is equal to the product of the volume of the solution V and its density ρ:

m=ρV, then

The mass fraction of potassium chloride in solution is equal to:

Example 2

What volume of a solution of nitric acid with a mass fraction of HNO 3 30% (ρ=1180kg/m 3 ) is required to prepare 20 liters of a 0.5 M solution of this acid?

Solution: First, we determine the mass of nitric acid in 20 liters of a 0.5 M solution:

M (HNO 3 )=63.01 g/mol;

m HNO 3 u003d 0.5 ∙ 63.01 ∙ 20 u003d 630.1 g.

Let’s determine in what volume of a solution with a mass fraction of HNO 3 30% contains 630.1 g of HNO 3 :

Therefore, in order to prepare 20 liters of 0.5 M HNO 3 , it is necessary to use only 1.78 liters of a solution of nitric acid with a mass fraction of HNO 3 equal to 30%.

Example 3

What mass of a solution with a mass fraction of KOH of 20% must be added to 250 g of a solution with a mass fraction of KOH of 90% to obtain a solution with ω KOH = 50%?

Solution: The problem is solved using the mixing rule. The mass of the solution with a mass fraction of KOH 20% is denoted by x.

Then 3x=1000; x=333.3.

To obtain a solution with a mass fraction of KOH of 50%, it is necessary to add 333.3 g of a KOH solution with 20% to 250 g of a KOH solution with ω = 90%.

Problems of this type are solved using a diagonal scheme or the “rule of the cross”: the point of intersection of two line segments denotes the properties of the mixture to be obtained.

20 (90-50)=40

fifty

90 (50-20)=30

The masses of the initial solutions required to prepare the mixture are inversely proportional to the differences between the concentrations of a given and less concentrated solution and more concentrated and given solutions:

Also, this problem can be solved, given that when two solutions are drained, the mass of the solute is summed up. Let the mass of a 20% solution be x g, then the mass of KOH in it is 0.2 x. The mass of KOH in the second solution is 0.9 × 250 = 225 g. The mass of the substance in the final solution is 0.5 × (250 + x). Thus,

0.2x + 225 = 0.5(250+x); x=333.3 g.

1.2. MOLAR AND EQUIVALENT CONCENTRATIONS

Example 1

What is the mass of NaOH contained in 0.2 l of solution if the molar concentration of the solution is 0.2 mol/l?

Decision:

Molar concentration C m or M (molarity) shows the amount of solute contained in 1 liter of solution.

Molar concentration (mol / l) is expressed by the formula

where m 1 is the mass of the dissolved substance, g;

M is the molar mass of the solute, g/mol;

V is the volume of the solution, l.

M (NaOH)=40 g/mol. The mass of NaOH contained in the solution is

M NaOH =MV=0.2∙40∙0.2=1.6 g.

Example 2

Determine the molar concentration of the equivalent of iron chloride (ІІІ), if 0.3 l of the solution contains 32.44 g of FeCl 3 .

Decision:

The molar concentration of a substance equivalent (normality) shows the number of molar mass equivalents of a solute contained in 1 liter of solution (mol/l):

where m A is the mass of the dissolved substance, g;

M (1/zA) is the molar mass of the solute equivalent, g/mol;

V is the volume of the solution, l.

The molar mass of the FeCl 3 equivalent is

The molar concentration of the equivalent solution of FeCl 3 is

Example 3

Determine the concentration of the KOH solution, if neutralization is 0.035 l 0.3 N. H 3 PO 4 consumed 0.02 l of KOH solution.

Decision:

From the law of equivalents it follows that the number of equivalents of substances involved in a chemical reaction is the same. 0.035 0.3 = 0.0105 equivalents of phosphoric acid are involved in the reaction. To neutralize H 3 PO 4

the same amount of substance equivalent to KOH will be required, i.e.

V (H 3 PO 4 ) C H (H 3 PO 4 ) u003d V (KOH) C H (KOH).

From here

1.3. MOLAL CONCENTRATION (MOLALITY), MOLAL FRACTION, TITER

Example 1

In what mass of ether must 3.04 g of aniline C 6 H 5 NH 2 be dissolved in order to obtain a solution whose molality is 0.3 mol / kg?

Decision:

The molality of a solution Cm (mol / kg) shows the amount of a solute in 1 kg of a solvent:

where m p-la is the mass of the solvent, kg;

n (A) is the amount of solute, mol.

M (C 6 H 5 NH 2 ) – 99.13 g/mol.

The mass of the solvent (ether) is equal to:

then

Example 2

Determine the titer of 0.01 N. CON.

Decision:

The titer of a solution indicates the mass (g) of the solute contained in 1 ml of the solution. In 1 l 0.01 n. KOH contains 0.564 g KOH. The titer of this solution is:

T= 0.561/1000=0.000561 g/ml.

Example 3

Calculate the molar fractions of glucose C 6 H 12 O 6 and water in a solution with a mass fraction of glucose of 36%.

Decision:

The mole fraction of a substance A (χ A ) in a solution is equal to the ratio of the amount of this substance n A to the total amount of all substances contained in the solution:

where ( ) the amount of all substances contained in the solution.

100 g of a solution with a mass fraction of glucose equal to 36% contains 36 g of glucose and 64 g of water:

nC 6 H 12 O 6 u003d 36/180 u003d 0.20 mol;

nH 2 O u003d 64/18 u003d 3.56 mol;

nC 6 H 12 O 6 + nH 2 O u003d 0.20 + 3.56 u003d 3.76 mol;

χC 6 H 12 O 6 u003d 0.20 / 3.76 u003d 0.053;

χH 2 O u003d 3.56 / 3.76 u003d 0.947.

The sum of the mole fractions of all components of the solution is 1.

Example 4

Calculate the molar concentration of the equivalent, the molar concentration and the molality of the solution in which the mass fraction of CuSO 4 is 10%. The density of the solution is 1107 kg/m 3 .

Decision:

Let’s determine the molar mass and the molar mass of the CuSO 4 equivalent:

M (CuSO 4 )= 159.61 g/mol; M(1/2 CuSO 4 )=

100 g of a solution with ωCuSO 4 = 10% contains 10.0 g CuSO 4 and 90 g H 2 O.

Therefore, the molality of a CuSO 4 solution is

Сm (CuSO 4 /H 2 O) u003d 10 / (159.61 ∙ 0.09) u003d 0.696 mol / kg.

Molar concentration and molar equivalent concentration refer to 1 liter of solution:

m solution u003d ρV u003d 1107 10 -3 u003d 1.107 kg.

This mass of solution contains 1.107 0.1 = 0.1107 kg of CuSO 4, which is 110.7 / 159.61 = 0.693 mol, or 0.693 2 = 1.386 eq.

The molar concentration and the molar concentration of the equivalent of this solution are 0.693 and 1.386 mol/L, respectively.

1.4. OSMOTIC PRESSURE. VANT HOFF’S LAW

Example 1

Calculation of the osmotic pressure of solutions.

Calculate the osmotic pressure of a solution containing

1.4 l 63 g glucose C 6 H 12 O 6 at 0°C.

Decision:

The osmotic pressure of the solution is determined according to the van’t Hoff law:

Pocm = nRT/V,

where n is the amount of solute, mol;

V is the volume of the solution, m 3 ;

R is the molar gas constant, equal to 8.3144 J/(mol-K).

1.4 l of solution contains 63 g of glucose, the molar mass of which is 180.16 g/mol. Therefore, 1.4 l of the solution contains

n= 63/180.16=0.35 mol of glucose.

The osmotic pressure of this glucose solution is:

Example 2

Determination of the molecular weight of non-electrolyte by the osmotic pressure of the solution.

Calculate the molecular weight of the non-electrolyte if 5 liters of the solution contains 2.5 g of the non-electrolyte. The osmotic pressure of this solution is 0.23∙10 5 Pa at 20°C.

Decision:

Replacing n with m/M, where m is the mass of the solute and M is its molar mass, we get

Rosm = mRT/(MV).

Hence the molar mass of the solute is

Therefore, the molecular weight of the non-electrolyte is 52.96

Rosm kPa: R=8.31 J/mol∙K;

Rosm mm Hg st.: R = 62.32 l ∙ mm Hg / deg ∙ mol;

Rosm. atm.: R=0.082 l∙atm../deg.∙mol.

1.5. PRESSURE OF SATURATED VAPOR OF SOLUTIONS. RAHOULT’S TONOMETRIC LAW

Example 1. a) Calculate the vapor pressure over a solution containing 34.23 g of sugar C 12 H 22 O 11 in 45.05 g of water at 65 ºС, if the vapor pressure of water at this temperature is 2.5 10 4 Pa.

Decision:

The vapor pressure of a solution of a non-volatile substance in a solvent is always lower than the vapor pressure of a pure solvent at the same temperature. The relative decrease in the vapor pressure of the solvent over the solution according to Raoult’s law is expressed by the relation

where p 0 is the vapor pressure over a pure solvent;

p is the vapor pressure of the solvent over the solution;

n is the amount of the dissolved substance, mol;

N is the amount of solvent, mol;

M (C 12 H 22 O 11 ) = 342.30 g/mol;

M (H 2 O) u003d 18.02 g / mol.

Amount of solute and solvent: n=34.23/342.30=0.1 mol; N u003d 45.05 / 18.02 u003d 2.5 mol.

Vapor pressure over the solution:

Example 2 Calculate the molecular weight of a non-electrolyte if 28.5 g of this substance dissolved in 785 g of water causes a decrease in the vapor pressure of water over the solution by 52.37 Pa at 40°C. The water vapor pressure at this temperature is 7375.9 Pa.

Decision:

The relative decrease in the vapor pressure of the solvent over the solution is

We find:

here m x is the mass of non-electrolyte, the molar mass of which is M x g/mol.

0.309M x + 0.202=28.5;

0.309Mx =28.298;

M x =91.58 g/mol.

The molecular weight of the non-electrolyte is ~ 92.

1.6. BOILING AND FREEZING TEMPERATURES OF SOLUTIONS.

EBUllIOSCOPIC AND CRYOSCOPIC RAOULT’S LAWS

Example 1. Determine the boiling and freezing point of a solution containing 1 g of nitrobenzene C 6 H 5 NO 2 in 10 g of benzene. Ebulioscopic and cryoscopic constants are 2.57 and 5.1 °C. The boiling point of pure benzene is 80.2 °C, the freezing point is -5.4 °C.

Decision:

According to Raoult’s law:

where ∆t zam and ∆t kip are, respectively, the decrease in the freezing point and the increase in the boiling point of the solution; K c and K e are, respectively, the cryoscopic and ebullioscopic constants of the solvent; g is the mass of the dissolved substance, g; G is the mass of the solvent, g; Mr is the molecular weight of the solute; Mr(C 6 H 5 NO 2 ) = 123.11.

Increasing the boiling point of a solution of nitrobenzene in benzene:

The boiling point of the solution: t bale u003d 80.2 + 2.09 u003d 82.29 ° C.

Lowering the freezing point of a solution of nitrobenzene in benzene:

The freezing point of the solution t deputy u003d 5.4 – 4.14 u003d 1.26 ° C.

Example 2. A solution of camphor weighing 0.522 g in 17 g of ether boils at a temperature 0.461 ° C higher than pure ether. The ebullioscopic constant of the ether is 2.16 ºС. Determine the molecular weight of camphor.

Decision:

The molecular weight of camphor is determined using the ratio

M r =

The molecular weight of camphor is 155.14.

Be First to Comment

Leave a Reply

Your email address will not be published.