Question

1) list the following aqueous solution in order of decreasing freezing point: 0.040 m glycerin ( C3H8O3), 0.020 m KBr, 0.030 m phenol (C6H5OH).

2) Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1°C. What isthe approximate molar mass of lauryl alcohol?

Answer 1

General guidance

Concepts and reason

Freezing point of a substance is the temperature at which it undergoes phase transition from liquid to solid. Freezing point can be determined by knowing number of particles present in a solution. Molar mass of any substance is calculated by summing individual atomic mass.

Fundamentals

• Freezing point depression: It a colligative property which depends on the number of particles present in a solution.

• $\\begin{array}{l}\\\\{\\rm{\\Delta }}{{\\rm{T}}_{\\rm{f}}}{\\rm{ = }}{{\\rm{K}}_{\\rm{f}}}{\\rm{m}}\\\\\\\\{\\rm{\\Delta }}{{\\rm{T}}_{\\rm{f}}}{\\rm{ = }}\\,{\\rm{change}}\\,{\\rm{in}}\\,{\\rm{freezing}}\\,{\\rm{point}}\\\\\\\\{{\\rm{K}}_{\\rm{f}}}\\,\\,{\\rm{ = }}\\,{\\rm{molal}}\\,{\\rm{freezing}}\\,{\\rm{depression}}\\,{\\rm{point}}\\,{\\rm{constant}}\\\\\\\\{\\rm{m}}\\,\\,\\,\\,{\\rm{ = }}\\,{\\rm{molality}}\\\\\\end{array}$

• ${\\rm{molality}}\\,{\\rm{ = }}\\frac{{{\\rm{moles of}}\\,{\\rm{solute}}\\,}}{{{\\rm{Kg}}\\,{\\rm{of}}\\,{\\rm{solvent}}}}$

• Electrolytes will have the highest freezing point and non-electrolytes will have a lower freezing point.

• ${\\rm{molar}}\\,{\\rm{mass}}\\,{\\rm{ = }}\\frac{{{\\rm{mass}}\\,}}{{{\\rm{moles}}\\,{\\rm{of}}\\,{\\rm{substance}}}}$

Step-by-step

Step 1 of 4

(1)

Electrolytes: KBr, Phenol

Non-electrolytes: Glycerin

Number of particles in an aqueous solution:

0.020 m KBr: 0.04 m

0.030 m Phenol: 0.06m

0.040 m glycerin: 0.040 m

Explanation | Hint for next step

KBr is strong electrolytes which can give ${\\rm{0}}{\\rm{.02 \\times 2 }}\\left( {{{\\rm{K}}^{\\rm{ + }}}{\\rm{,B}}{{\\rm{r}}^{\\rm{ - }}}} \\right){\\rm{ = 0}}{\\rm{.04 m of}}\\,{{\\rm{K}}^{\\rm{ + }}}{\\rm{and B}}{{\\rm{r}}^{\\rm{ - }}}$Phenol is strong electrolyte which can give ${\\rm{0}}{\\rm{.03 \\times 2}}\\left( {{\\rm{P}}{{\\rm{h}}^{\\rm{ - }}}{\\rm{,}}{{\\rm{H}}^{\\rm{ + }}}} \\right)\\,\\,{\\rm{ = }}\\,{\\rm{0}}{\\rm{.06 m of P}}{{\\rm{h}}^{\\rm{ - }}}{\\rm{\\;and }}{{\\rm{H}}^{\\rm{ + }}}$Glycerin is weak electrolytes which will not dissociate in an aqueous solution.

Step 2 of 4

${\\rm{Phenol }}\\left( {{\\rm{0}}{\\rm{.06 m}}} \\right){\\rm{ > Glycerin }}\\left( {{\\rm{0}}{\\rm{.04}}} \\right){\\rm{ = KBr }}\\left( {{\\rm{0}}{\\rm{.04}}} \\right)$

Part 1

Decreasing order of freezing point is given below:

${\\rm{Phenol > Glycerin = KBr }}$

Explanation | Common mistakes | Hint for next step

As we know from depression in freezing point equation change in freezing point of solution depends on the number of particle dissociated on solvation.

The order of the decrease in freezing point of the solution will correspond to the concertation of particles present in a solution

Step 3 of 4

(2)

$\\begin{array}{l}\\\\{\\rm{\\Delta }}{{\\rm{T}}_{\\rm{f}}}{\\rm{ = }}{{\\rm{K}}_{\\rm{f}}}{\\rm{m}}\\\\\\\\{\\rm{4}}{\\rm{.1^\\circ C = 4}}{\\rm{.90^\\circ C/m \\times }}\\,{\\rm{x}}\\,{\\rm{m}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( {{{\\rm{K}}_{\\rm{f}}}{\\rm{ of benzene = 4}}{\\rm{.90^\\circ C/m}}} \\right)\\\\\\\\{\\rm{x}}\\,{\\rm{ = 0}}{\\rm{.8367}}\\,{\\rm{m}}\\,\\\\\\end{array}$

Explanation | Common mistakes | Hint for next step

Molality of a chemical substance is calculated by substituting ${\\rm{\\Delta }}{{\\rm{T}}_{\\rm{f}}}\\,{\\rm{and}}\\,{{\\rm{K}}_{\\rm{f}}}\\,$values in depression in freezing point equation.

Step 4 of 4

$\\begin{array}{l}\\\\{\\rm{molality}}\\,{\\rm{ = }}\\frac{{{\\rm{moles of}}\\,{\\rm{solute}}\\,}}{{{\\rm{Kg}}\\,{\\rm{of}}\\,{\\rm{solvent}}}}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{\\rm{ = }}\\frac{{{\\rm{mass/molar}}\\,{\\rm{mass}}}}{{{\\rm{Kg}}\\,{\\rm{of}}\\,{\\rm{solvent}}}}\\\\\\\\{\\rm{0}}{\\rm{.273}}\\,{\\rm{mol}}\\,{\\rm{k}}{{\\rm{g}}^{{\\rm{ - 1}}}}{\\rm{ = }}\\frac{{{\\rm{5}}{\\rm{.00g/x\\;g}}\\,{\\rm{mo}}{{\\rm{l}}^{{\\rm{ - 1}}}}}}{{{\\rm{0}}{\\rm{.100}}\\,{\\rm{kg}}}}\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{\\rm{ x}}\\,{\\rm{ = }}\\,{\\rm{183}}{\\rm{.156}}\\,{\\rm{g mo}}{{\\rm{l}}^{{\\rm{ - 1}}\\,\\,}}\\,\\\\\\end{array}$

Part 2

Molar mass of lauryl alcohol is${\\rm{183}}{\\rm{.156}}\\,{\\rm{g mo}}{{\\rm{l}}^{{\\rm{ - 1}}\\,\\,}}$.

Molar mass of a chemical substance is calculated by dividing the mass by mole present in it.

Answer

Part 1

Decreasing order of freezing point is given below:

${\\rm{Phenol > Glycerin = KBr }}$

Part 2

Molar mass of lauryl alcohol is${\\rm{183}}{\\rm{.156}}\\,{\\rm{g mo}}{{\\rm{l}}^{{\\rm{ - 1}}\\,\\,}}$.

Answer only

Part 1

Decreasing order of freezing point is given below:

${\\rm{Phenol > Glycerin = KBr }}$

Part 2

Molar mass of lauryl alcohol is${\\rm{183}}{\\rm{.156}}\\,{\\rm{g mo}}{{\\rm{l}}^{{\\rm{ - 1}}\\,\\,}}$.