Name: Solid Solutions
Uploaded: 2019-07-19
Duration: 2 min 56 s
Description: Solutions can be categorized as solid, liquid, and gaseous based on the physical states of solvents and solutes. In this video, let's learn about solid solutions.

Question

A solution containing

0.5 g

of

KCl

dissolves in

100 g

of water and freezes at

- 0.24 ° C

. Calculate the degree of dissociation of the salt. (K_f for water

= 1.86 K \cdot Kg \cdot {mol}^{- 1}

) Atomic weights,

K = 39, Cl = 35.5

.

Accepted Answer

Freezing point of the given solution ( $T^{0}$ ) $=$ $- 0.24 ° C$

We know that freezing point of pure water ( $T_{f}$ ) = $0 ° C$

Depression in freezing point ( $∆ T_{f}$ ) is:

$T_{f} - T^{0} = 0 - (- 0.24) = 0.24 ° C$

[Note: The value of $∆ T_{f}$ is the same on both Celsius and Kelvin scale.]

Given,

Molal depression constant or cryoscopic constant ( $K_{f}$ ) $=$ $1.86 K \cdot Kg \cdot {mol}^{- 1}$

Molality of the solution ( $m$ ) is given by the following formula:

$Molality of the solution (m) = \frac{No' of moles of solute}{Weight of solvent, water (in Kg)}$

Number of moles of solute = $\frac{mass of solute in grams}{molecular mass of solute in g \cdot {mol}^{- 1}}$

[ molecular mass of $KCl$ $= 39 + 35.5 = 74.5 g / mol$ , mass of $KCl$ (given) $= 0.5 g$ , and weight of water (in Kg) (given) $= 100 g = 0.1 Kg$ ]

$∴ m = \frac{\frac{0.5}{74.5}}{0.1} = \frac{0.5}{74.5 \times 0.1} = 0.067114094 \approx 0.067$

The van't Hoff equation for depression of freezing point, where, i is the van't of factor, is given by:

$∆ T_{f} = i \cdot K_{f} \cdot m 0.24 = i \times 1.86 \times 0.067 \Rightarrow i = \frac{0.24}{1.86 \times 0.067} = 1.925854598 \approx 1.93$

Potassium chloride dissociates in solution as follows. Suppose, we start with one mole of potassium chloride and $α$ is its degree of dissociation.

$KCl ⇌ K^{+} + {Cl}^{-} Initially 1 mole______At equilibrium (1 - α) moles α moles α moles$

Total number of moles at equilibrium $= 1 - α + α + α = 1 + α$

Therefore, van't Hoff factor for potassium chloride, $i = \frac{No . of moles in solution}{No . of moles taken} = \frac{1 + α}{1} = 1 + α = 1.93 \Rightarrow α = 1.93 - 1 = 0.93 or 93 %$

Answer: The degree of dissociation of potassium chloride is $0.93 or 93 %$ .

A solution containing $0.5 g$ of $KCl$ dissolves in $100 g$ of water and freezes at $- 0.24 ° C$ . Calculate the degree of dissociation of the salt. (K_f for water $= 1.86 K \cdot Kg \cdot {mol}^{- 1}$ ) Atomic weights, $K = 39, Cl = 35.5$ .

Important Questions on Solutions

Learn and Prepare for any exam you want !

A solution containing 0.5 g of KCl dissolves in 100 g of water and freezes at -0.24°C. Calculate the degree of dissociation of the salt. (Kf for water =1.86 K·Kg·mol-1 ) Atomic weights, K = 39, Cl= 35.5.

Important Questions on Solutions

Learn and Prepare for any exam you want !

A solution containing $0.5 g$ of $KCl$ dissolves in $100 g$ of water and freezes at $- 0.24 ° C$ . Calculate the degree of dissociation of the salt. (K_f for water $= 1.86 K \cdot Kg \cdot {mol}^{- 1}$ ) Atomic weights, $K = 39, Cl = 35.5$ .