What is the boiling point of a solution formed by dissolving 0.75 mol of KCl in 1.00 kg of water?

Q1:

M = 1.0 mol/L.

Explanation:

•To solve this problem, we can use the relation:

M = 10Pd / Molar mass, where

M is the molarity of the solution (mol/L),

P is the percent of the solution (P = 7.06 %),

d is the density of the solution (d = 1.19 g/ml),

Molar mass of sodium bicarbonate (84.007 g/mol),

M = 10 (7.06 %) (1.19 g/ml) / (84.007 g/mol) = 1.0 mol/L.

Q2:

Molality = 0.904 m.

Explanation:

•To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

The solute is sodium bicarbonate (7.06 %), it is the component with the lower percent.

The solvent is water (92.94 %), it is the component with the higher percent.

•To get the mass of the solute and solvent, we should firstly get the mass of the solution.

Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (1.19 g/ml) (1000 ml) = 1190 g.

The mass of solute sodium bicarbonate = (the total mass of the solution x percent % of solute) / 100 = (1190 x 7.06) / 100 = 84.014 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (1190 x 92.94) / 100 = 1105.986 g.

m = (mass / molar mass) solute x (1000 / mass of solvent)

m = (84.014 g / 84.007 g/mol) (1000 / 1105.986 g) = 0.904 m.

Q3:  

Mole fraction of solvent (water) = 0.953.

Explanation:

•The solute is methanol CH3OH (8.05 %), it is the component with the lower percent.

•The solvent is water (91.95 %), it is the component with the higher percent.

•To solve this problem, we can use the relation of mole fraction of water (X water):

X water = (no. of moles of water) / (total no. of moles of the solution)

•To get the no. of moles of the solute and solvent, we should obtain the mass of both.

•To obtain the mass of the solute and solvent, we should firstly get the mass of the solution.

•Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (0.976 g/ml) (1000 ml) = 976.0 g.

The mass of solute methanol CH3OH = (the total mass of the solution x percent % of solute) / 100 = (976.0 x 8.05) / 100 = 78.568 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (976.0 x 91.95) / 100 = 897.432 g.

The no. of moles of solute methanol CH3OH = mass / molar mass = (78.568 g) / (32.0 g/mol) = 2.45525 mol.

The no. of moles of solvent water = mass / molar mass = (897.432 g) / (18.0 g/mol) = 49.857 mol.

X water = (no. of moles of water) / (total no. of moles of the solution) = (49.857) / (49.857 + 2.45525) = 0.953.

Q4:

m = 5.18 m.

Explanation:

•To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

Mass of the solvent water = 60.0 g.

m = (mass / molar mass) cesium chloride x (1000 / mass of water)

m = (52.3 g / 168.36 g/mol) (1000 / 60.0 g) = 5.177 m = 5.18 m.

Q5:

4.91 M.

Explanation:

•To solve this problem, we can use the relation of molarity (M):

M = (mass / molar mass) solute x (1000 / V of the solution)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

V of the solution = 63.3 ml.

M = (mass / molar mass) cesium chloride x (1000 / V of the solution)

M = (52.3 g / 168.36 g/mol) (1000 / 63.3 g) = 4.907 M = 4.91 M.

Q6:

The vapor pressure of the solution = 80.0 torr.

Explanation:

•The vapor pressure of the solution will be the partial vapor pressure of the ethanol only because alpha naphthol is assumed to be nonvolatile, so it has no partial vapor pressure in the total vapor pressure of the solution.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol),  

Where, X ethanol is the mole fraction of ethanol.

P0 ethanol is the vapor pressure of pure ethanol (P0 = 100.0 torr).

•To get the mole fraction of ethanol, we can use the relation of mole fraction of ethanol (X ethanol):

X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution).

•The no. of moles of ethanol = mass / molar mass = (36.8 g) / (46.07 g/mol) = 0.80 mol.

•The no. of moles of alpha naphthol = mass / molar mass = (28.8 g) / (144.0 g/mol) = 0.2 mol.

•X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution) = (0.80) / (0.80 + 0.2) = 0.80.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol) = (0.80) (100.0 torr) = 80.0 torr.

Kindly find the answers of all questions in details attached as a word and pdf file due to the limitations of the capacity of the answer.

Mass of solute, W_B = 8.05 g[/tex]

Molar mass of solute, M_B = 32 g /mol

Mass of solvent,

Molar mass of solvent,

Number of mol of solvent is

Number of mol of solute is

n_B = W_B /M_B

= 8.05 / 32 = 0.25 mol

Mole fraction of solvent is the number of moe of solvent divided by total nmber of mole of solute and solvent as follows:

Thus, mole fraction of solvent is 0.954

Q1:

M = 1.0 mol/L.

Explanation:

•To solve this problem, we can use the relation:

M = 10Pd / Molar mass, where

M is the molarity of the solution (mol/L),

P is the percent of the solution (P = 7.06 %),

d is the density of the solution (d = 1.19 g/ml),

Molar mass of sodium bicarbonate (84.007 g/mol),

M = 10 (7.06 %) (1.19 g/ml) / (84.007 g/mol) = 1.0 mol/L.

Q2:

Molality = 0.904 m.

Explanation:

•To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

The solute is sodium bicarbonate (7.06 %), it is the component with the lower percent.

The solvent is water (92.94 %), it is the component with the higher percent.

•To get the mass of the solute and solvent, we should firstly get the mass of the solution.

Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (1.19 g/ml) (1000 ml) = 1190 g.

The mass of solute sodium bicarbonate = (the total mass of the solution x percent % of solute) / 100 = (1190 x 7.06) / 100 = 84.014 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (1190 x 92.94) / 100 = 1105.986 g.

m = (mass / molar mass) solute x (1000 / mass of solvent)

m = (84.014 g / 84.007 g/mol) (1000 / 1105.986 g) = 0.904 m.

Q3:  

Mole fraction of solvent (water) = 0.953.

Explanation:

•The solute is methanol CH3OH (8.05 %), it is the component with the lower percent.

•The solvent is water (91.95 %), it is the component with the higher percent.

•To solve this problem, we can use the relation of mole fraction of water (X water):

X water = (no. of moles of water) / (total no. of moles of the solution)

•To get the no. of moles of the solute and solvent, we should obtain the mass of both.

•To obtain the mass of the solute and solvent, we should firstly get the mass of the solution.

•Let assume that the solution has a volume of 1.0 L (1000 ml).

Then the mass of the solution = d x V = (0.976 g/ml) (1000 ml) = 976.0 g.

The mass of solute methanol CH3OH = (the total mass of the solution x percent % of solute) / 100 = (976.0 x 8.05) / 100 = 78.568 g.

The mass of solvent water = (the total mass of the solution x percent % of solvent) / 100 = (976.0 x 91.95) / 100 = 897.432 g.

The no. of moles of solute methanol CH3OH = mass / molar mass = (78.568 g) / (32.0 g/mol) = 2.45525 mol.

The no. of moles of solvent water = mass / molar mass = (897.432 g) / (18.0 g/mol) = 49.857 mol.

X water = (no. of moles of water) / (total no. of moles of the solution) = (49.857) / (49.857 + 2.45525) = 0.953.

Q4:

m = 5.18 m.

Explanation:

•To solve this problem, we can use the relation of molality (m):

m = (mass / molar mass) solute x (1000 / mass of solvent)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

Mass of the solvent water = 60.0 g.

m = (mass / molar mass) cesium chloride x (1000 / mass of water)

m = (52.3 g / 168.36 g/mol) (1000 / 60.0 g) = 5.177 m = 5.18 m.

Q5:

4.91 M.

Explanation:

•To solve this problem, we can use the relation of molarity (M):

M = (mass / molar mass) solute x (1000 / V of the solution)

Mass of the solute cesium chloride = 52.3 g.

Molar mass of cesium chloride = 168.36 g/mol.

V of the solution = 63.3 ml.

M = (mass / molar mass) cesium chloride x (1000 / V of the solution)

M = (52.3 g / 168.36 g/mol) (1000 / 63.3 g) = 4.907 M = 4.91 M.

Q6:

The vapor pressure of the solution = 80.0 torr.

Explanation:

•The vapor pressure of the solution will be the partial vapor pressure of the ethanol only because alpha naphthol is assumed to be nonvolatile, so it has no partial vapor pressure in the total vapor pressure of the solution.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol),  

Where, X ethanol is the mole fraction of ethanol.

P0 ethanol is the vapor pressure of pure ethanol (P0 = 100.0 torr).

•To get the mole fraction of ethanol, we can use the relation of mole fraction of ethanol (X ethanol):

X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution).

•The no. of moles of ethanol = mass / molar mass = (36.8 g) / (46.07 g/mol) = 0.80 mol.

•The no. of moles of alpha naphthol = mass / molar mass = (28.8 g) / (144.0 g/mol) = 0.2 mol.

•X ethanol = (no. of moles of ethanol) / (total no. of moles of the solution) = (0.80) / (0.80 + 0.2) = 0.80.

The vapor pressure of the solution = the partial vapor pressure of ethanol = (X ethanol) (Po ethanol) = (0.80) (100.0 torr) = 80.0 torr.

Kindly find the answers of all questions in details attached as a word and pdf file due to the limitations of the capacity of the answer.

Mass of solute, W_B = 8.05 g[/tex]

Molar mass of solute, M_B = 32 g /mol

Mass of solvent,

Molar mass of solvent,

Number of mol of solvent is

Number of mol of solute is

n_B = W_B /M_B

= 8.05 / 32 = 0.25 mol

Mole fraction of solvent is the number of moe of solvent divided by total nmber of mole of solute and solvent as follows:

Thus, mole fraction of solvent is 0.954

- B) 79.1 degree C

Solution:- An elevation in boiling point takes place when a non volatile solute is added to the solvent and the equation used for this is:

where, is the elevation in boiling point

i is Van't hoff factor. alpha naphthol is covalent compound and so the value of i is 1.

m is molality of the solution and kb is the molal elevation constant for the solvent and it's value is given as 1.22

Molar mass of alpha-naphthol = 10(12)+8(1)+1(16)

= 120 + 8 + 16

= 144 gram per mole

let's calculate the moles of solute:

= 0.0556 mole

molality is moles of solute per kg of solvent. 0.0556 moles of solute are dissolved in 100.0 g that is 0.100 kg of ethanol.

So, molality =

molality = 0.556m

Let's plug in the values in the formula and solve for boiling point elevation:

= 0.678

Boiling point of the solution = 78.4 + 0.678 = 79.078 degree C

If we round this two one decimal place then it becomes 79.1 degree C.  Hence, choice B) 79.1 degree C is the right answer.

- B) 79.1 degree C

Solution:- An elevation in boiling point takes place when a non volatile solute is added to the solvent and the equation used for this is:

where, is the elevation in boiling point

i is Van't hoff factor. alpha naphthol is covalent compound and so the value of i is 1.

m is molality of the solution and kb is the molal elevation constant for the solvent and it's value is given as 1.22

Molar mass of alpha-naphthol = 10(12)+8(1)+1(16)

= 120 + 8 + 16

= 144 gram per mole

let's calculate the moles of solute:

= 0.0556 mole

molality is moles of solute per kg of solvent. 0.0556 moles of solute are dissolved in 100.0 g that is 0.100 kg of ethanol.

So, molality =

molality = 0.556m

Let's plug in the values in the formula and solve for boiling point elevation:

= 0.678

Boiling point of the solution = 78.4 + 0.678 = 79.078 degree C

If we round this two one decimal place then it becomes 79.1 degree C.  Hence, choice B) 79.1 degree C is the right answer.

  MCl₂

Explanation:

The formula for boiling point elevation can be used to find x. The "complete dissociation" means there will be an ion of M and x ions of Cl in the solution. The number of moles of solute will be 30.2 grams divided by the molecular weight of MClx, where x is the variable we're trying to find.

Then the formula for the salt is MCl₂.

Formula for the salt: MCl₃

Explanation:

MClₓ → M⁺  +  xCl⁻

We apply the colligative property of boiliing point elevation.

We convert the boiling T° to °C

375.93 K - 273K = 102.93°C

ΔT = Kb . m . i

where ΔT means the difference of temperature, Keb, the ebulloscopic constant for water, m the molality of solution (mol of solute/kg of solvent) and i, the Van't Hoff factor (numbers of ions dissolved)

ΔT = 102.93°C - 100°C = 2.93°C

Kb = 0.512 °C/m

We replace data: 2.93°C = 0.512 °C/m . m . i

i = x + 1 (according to the equation)

22.9 g / (56g/m + 35.45x) = moles of salt / 0.1kg = molality

We have calculated the moles of salt in order to determine the molar mass, cause we do not have the data. We replace

2.93°C = 0.512 °C/m . [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

2.93°C / 0.512 m/°C = [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

5.72 m = [22.9 g / (56g/m + 35.45x)]/ 0.1 (x+1)

5.72 . 0.1 / [22.9 g / (56g/m + 35.45x)] = x+1

0.572 / (22.9 g / (56g/m + 35.45x) = x+1

0.572 (56 + 35.45x) / 22.9 = x+1

0.572 (56 + 35.45x) = 22.9x + 22.9

32.03 + 20.27x = 22.9x + 22.9

9.13 = 2.62x

x = 3.48 ≅ 3

  MCl₂

Explanation:

The formula for boiling point elevation can be used to find x. The "complete dissociation" means there will be an ion of M and x ions of Cl in the solution. The number of moles of solute will be 30.2 grams divided by the molecular weight of MClx, where x is the variable we're trying to find.

Then the formula for the salt is MCl₂.