# Solution

## Elevation of Boiling Point

Boiling point of a solution is always higher than that of the pure solvent. For dilute solutions, elevation of boiling point is directly proportional to the molar concentration of solute in a solution.

`ΔT_b∝m`

Or, `ΔT_b=K_bm`

Here, m is molality and K_{b} is called **Boiling Point Elevation Constant** or **Molal Elevation Constant (Ebullioscopic Constant)**. The unit of K_{b} is K kg mol^{-1}

If w_{2} g of solute of molar mass M_{2} is dissolved in w_{1} g of solvent. Then molality of solution is given as follows:

`m=((w_2)/(M_2))/((w_1)/(1000))-(1000xx\w_2)/(M_2xx\w_1)`

Substituting the value of molality in previous equation we get:

`ΔT_b=(K_b\xx1000xx\w_2)/(M_2xx\w_1)`

Or, `M_2=(1000xx\w_2xx\K_b)/(ΔT_b\xx\w_1)`

**Example:** 18 g of glucose C_{6}H_{12}O_{6} is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? K_{b} for water is 0.52 K kg mol^{-1}.

**Answer:** Moles of glucose `=(18g)/(180g\text(mol)^(-1))=0.1` mol

Mass of solvent = 1 kg

Hence, molality of glucose solutin = 0.1 mol kg^{-1}

For water, change in boiling point

Δ `T_b=K_b\xx\m`

`=0.52K\kg\text(mol)^(-1)xx0.1text(mol)kg^(-1)=0.052` K

Boiling point of water at 1.013 bar is 373.15 K

So, boiling point of solution `= 373.15 + 0.052 = 373.202` K

**Example:** The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_{b} for benzene is 2.53 K kg mol^{-1}.

**Answer:** ΔT_{b} = 354.11 K – 353.23 K = 0.88 K

So, molar mass of solute can be calculated as follows:

`M_2=(1000xx\w_2xx\K_b)/(ΔT_b\xx\w_1)`

`=(1000xx1.8xx2.53)/(0.88xx90)=57.5` g mol^{-1}

## Depression of Freezing Point

When a non-volatile solid is added to the solvent, its vapor pressure decreases and becomes equal to that of solid solvent at lower temperature. This results in lowering of freezing point of the solvent.

Depression of freezing point for dilute solute is directly proportional to molality.

`ΔT_f∝m`

Or, `ΔT_t=K_t\m` ………(1)

Here, the constant of proportionality, K_{1}, is called **Freezing Point Depression Constant**, **Depression Constnat** or **Cryoscopic Constant**. The unit of K_{f} is K kg mol^{-1}

We know that molality is given by following equation:

`m=((w_2)/(M_2))/((w_1)/(1000))`

So, the equation for ΔT_{f} can be written as follows:

`ΔT_f=(K_f\xx\w_2xx1000)/(ΔT_f\xx\w_1)` ……… (2)

**Example:** 45 g of ethylene glucol (C_{2}H_{6}O_{2} is mixed with 600 g of water. Calculate (a)the freezing point depression and (b) the freezing point of the solution.

**Answer:** First of all, we need to find molality of ethylene glycol.

Moles of ethylene glycol `=(45g)/(62g\text(mol)^-1)=0.73` mol

Mass of water `=(600g)/(1000kg)=0.6` kg

So, molality of ethylene glycol `=(0.73text(mol))/(0.60 kg)=1.2` mol kg^{-1}

Now, change in freezing point can be calculated as follows:

`ΔT_f=1.86kg\text(mol)^(-1)xx1.2 text(mol)kg^(-1)=2.2` K

So, freezing point of aqueous solution `= 273.15 K – 2.2 K = 270.95` K

**Example:** 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^{-1}. Find the molar mass of the solute.

**>Answer:** Molar mass of solute can be calculated as follows:

`M_2=(5.12 kg\text(mol)^(-1)xx1.00g\xx1000\g\kg^(-1))/(0.40xx50g)=256` g mol^{-1}