Question

Table salt has a Ksp in water of 37.9 M2 at 25°C. What will be the maximum boiling temperature of an aqueous sodium chloride solution made by saturating a solution(establishing a dynamic equilibrium of the salt and ions in solution)at 25°C, filtering off any solid remaining, then bringing the solution to a boil.

a)Find the molality (include units) for sodium chloride in a saturated aqueous solution:

b)Find the boiling point elevation and boiling point of the saturated solution. (If you were unable to calculate the molality in part (a), use m(NaCl) = 5 for part (b).

c)What will be the vapor pressure of water (in torr) above this solution at 25°C?

Answer 1

The dissolution of table salt in water is represented as

NaCl(s) Na⁺(aq) +Cl^-(aq)

Ksp = [Na⁺(aq) ][Cl^-(aq)]

Given

Ksp= 37.5 M²

Let, [Na⁺(aq) ]=[Cl^-(aq)]=s , at equilibrium.

Ksp = [Na⁺(aq) ][Cl^-(aq)]

37.5 M² = s.s

37.5 M² =s²

s=6.16 M

Therefore, at equilibrium  [Na⁺(aq) ]=[Cl^-(aq)]=6.16 M

Since NaCl is a strong electrolyte, most the NaCl will be dissociated.

Therefore, Molarity of NaCl solution =6.16 M = 6.16 mol/L

Density of saturated  NaCl solution at 25°C= 1.00409 g/ mL [Literature value]

Density= mass/volume

Mass of 1 L (1000mL) of saturated  NaCl solution at 25°C=  1.00409 g/ mL * 1000 mL

=1004.09 g

=1.00409 kg

Molarity of NaCl solution =6.16 M = 6.16 mol/L

Therefore, No. of moles of NaCl in 1L solution =6.16 moles

Molar mass of NaCl =58.44 g/mol

Mass of 6.16 moles of NaCl =58.44 g/mol * 6.16 moles

= 359.99 g

Mass of NaCl in 1L solution = 359.99 g = 0.3599 kg

Mass of 1 L of saturated  NaCl solution = 1.00409 g

Therefore,Mass of NaCl in 1.00409 Kg = 0.3599 kg

Mass of solvent (water) = Mass of solution - mass of solute

=  1.00409 kg -0.3599 kg

= 0.64419 kg

Molality(m) of solution= No. of moles of solute/ mass of solvent inkg

=6.16 moles/ 0.64419 kg

= 9.562 mol/ kg

Using the colligative property of elevation in boiling point

T_b = i K_b m

T_b = elevation in b.p = T_b - T_b^o , where T_b is the boiling point of solution and T_b^o is the boiling point of pure solvent.

i= Van't hoff factor, for NaCl i=2, as it is a strong electrolyte and almost completely dissociated.

K_b = Ebullioscopic constt,for water the K_b=0.512 (K·kg)/mol

m= molality of solution =  9.562 mol/ kg

T_b - T_b^o = i K_b m

T_b - 273.15 K = 2 * 0.512 (K·kg)/mol *  9.562 mol/ kg

T_b=282.94 K   

The maximum boiling temperature of saturated solution of table salt=282.94 K   

a)

The dissolution of table salt in water is represented as

NaCl(s) Na⁺(aq) +Cl^-(aq)

Ksp = [Na⁺(aq) ][Cl^-(aq)]

Given

Ksp= 37.5 M²

Let, [Na⁺(aq) ]=[Cl^-(aq)]=s , at equilibrium.

Ksp = [Na⁺(aq) ][Cl^-(aq)]

37.5 M² = s.s

37.5 M² =s²

s=6.16 M

Therefore, at equilibrium  [Na⁺(aq) ]=[Cl^-(aq)]=6.16 M

Since NaCl is a strong electrolyte, most the NaCl will be dissociated.

Therefore, Molarity of NaCl solution =6.16 M = 6.16 mol/L

Density of saturated  NaCl solution at 25°C= 1.00409 g/ mL [Literature value]

Density= mass/volume

Mass of 1 L (1000mL) of saturated  NaCl solution at 25°C=  1.00409 g/ mL * 1000 mL

=1004.09 g

=1.00409 kg

Molarity of NaCl solution =6.16 M = 6.16 mol/L

Therefore, No. of moles of NaCl in 1L solution =6.16 moles

Molar mass of NaCl =58.44 g/mol

Mass of 6.16 moles of NaCl =58.44 g/mol * 6.16 moles

= 359.99 g

Mass of NaCl in 1L solution = 359.99 g = 0.3599 kg

Mass of 1 L of saturated  NaCl solution = 1.00409 g

Therefore,Mass of NaCl in 1.00409 Kg = 0.3599 kg

Mass of solvent (water) = Mass of solution - mass of solute

=  1.00409 kg -0.3599 kg

= 0.64419 kg

Molality(m) of solution= No. of moles of solute/ mass of solvent inkg

=6.16 moles/ 0.64419 kg

= 9.562 mol/ kg

b)

Using the colligative property of elevation in boiling point

T_b = i K_b m

T_b = elevation in b.p = T_b - T_b^o , where T_b is the boiling point of solution and T_b^o is the boiling point of pure solvent.

i= Van't hoff factor, for NaCl i=2, as it is a strong electrolyte and almost completely dissociated.

K_b = Ebullioscopic constt,for water the K_b=0.512 (K·kg)/mol

m= molality of solution =  9.562 mol/ kg

T_b - T_b^o = i K_b m

T_b - 273.15 K = 2 * 0.512 (K·kg)/mol *  9.562 mol/ kg

T_b=282.94 K

Boiling point of saturated solution = 282.94 K

Elevation in boiling point = 282.94K - 273.15K =9.79 K

c) Vapour pressure of pure water at 25^oC = 23.8 torr [ litrature value]

Using Rauolt's law,

p_H2O  =  p^o_H2O

p_H2O = vapour pressure of water in solution

p^o_H2O=vapour pressure of pure water

= Mole fraction of water in solution

Calculation of mole fraction of water

Mass of water in 1L solution = 0.64419 kg = 644.19 g [already done in part (a)]

Molar mass of water = 18 g/mol

No. of moles of water in 1L solution = 644.19 g / ( 18 g/mol ) = 35.79 moles

Moles of NaCl in1L solution = 6.16 moles

Mole fraction of water in solution = No. of moles of water / [No. of moles of water + No. of moles of NaCl]

= 35.79/[35.79 + 6.16]

=0.853

Calculation of vapour pressure of water in solution

p_H2O  =  p^o_H2O

= 23.8 torr * 0.853

=20.3 torr