Question

Calculate the molal elevation constant of water, it is given that the 0.1 molal aqueous solutions of the substance boiled at 100.052 degree Celsius

Answer 1

Answer:

Delta Tb = Kb × m

100.052 = Kb × 0.1

100.052/ 0.1 = Kb

Kb = 1000.52

Answer 2

Answer:

The water's molal elevation constant measured is $0.52\textdegree C molal^{-1}$.

Explanation:

Given,

The molality of the aqueous solution = $0.1 molal$.

The boiling temperature, $T_{b}$ = $100.052 \textdegree C$

The water's molal elevation constant =?

From the relation given below, we can find out the molal elevation constant:

• $\Delta T = K_{b} \times m$   -------equation (1)

Here,

• ΔT = The change in the boiling point
• $K_{b}$ = The molal elevation constant
• m = Molality

Firstly. we have to calculate the change in the boiling point, ΔT.

• $\Delta T = T_ {b}- T$

Here,

• $T_{b}$ = The given boiling temperature
• T = The boiling temperature of water = $100\textdegree C$

Therefore,

• $\Delta T = 100.052 -100$ = $0.052\textdegree C$

Now, after putting the value of ΔT in equation (1), we get:

• $0.052 = K_{b} \times 0.1$
• $k_{b} = 0.52\textdegree C molal^{-1}$

Hence, the water's molal elevation constant, $k_{b} = 0.52\textdegree C molal^{-1}$.