Q.The unit of ebullioscopic constant is _______________.

The key idea is that the ebullioscopic constant ($K_b$) relates the boiling point elevation to molality: $\Delta T_b = K_b \cdot m$. Since $\Delta T_b$ is in Kelvin (K) and molality $m$ is in $\text{mol kg}^{-1}$, solving for $K_b$ gives units of $\frac{\text{K}}{\text{mol kg}^{-1}} = \text{K kg mol}^{-1}$. …

The ebullioscopic constant (also called the boiling point elevation constant) appears in the formula for boiling point elevation:

Here, $\Delta T_b$ is the elevation in boiling point (measured in Kelvin, K), and $m$ is the molality of the solution (measured in mol/kg, i.e., moles of solute per kilogram of solvent). The constant $K_b$ is a property of the solvent, not the solute.

To find the unit of $K_b$, we simply rearrange the equation:

Now substitute the units:

• $\Delta T_b$ has unit: K (Kelvin)
• $m$ has unit: mol/kg (or mol kg$^{-1}$)

So:

That is: $K\ kg\ mol^{-1}$.

1. Why not the other options?

   Option (ii) gives $mol\ kg\ K^{-1}$ — that would be the reciprocal of $K_b$, i.e., $1/K_b$.

   Option (iii) gives $kg\ mol^{-1}\ K^{-1}$ — that has an extra $K^{-1}$ in the denominator, which would make $K_b$ dimensionless when multiplied by molality, which is wrong.

   Option (iv) gives $K\ mol\ kg^{-1}$ — that is $K \cdot$ (molality), which is actually the unit of $\Delta T_b$, not $K_b$.

2. A quick check using the formula

   If $K_b$ had units of $K\ kg\ mol^{-1}$, then $K_b \times m$ gives: …

Concept: Ebullioscopic Constant (Boiling Point Elevation Constant)

The ebullioscopic constant ($K_b$) relates the elevation in boiling point to the molality of the solution.

Formula to remember

Where:

• $\Delta T_b$ = elevation in boiling point (unit: K)
• $m$ = molality of solution (unit: mol/kg)

Method: Dimensional Analysis

Steps

1. Write the defining equation

2. Substitute the units

   • $\Delta T_b$ has unit: K (kelvin)
   • $m$ has unit: mol/kg (moles per kilogram)

3. Perform the division …

Common Mistakes on the Unit of Ebullioscopic Constant ($K_b$)

The Correct Answer

The ebullioscopic constant $K_b$ has units of $K\ kg\ mol^{-1}$ or equivalently $K\ (molality)^{-1}$.

So option (i) is correct.

Mistake #1: Confusing $K_b$ with its reciprocal or cryoscopic constant

What students do wrong:

They pick option (iii) — $kg\ mol^{-1}\ K^{-1}$ — thinking $K_b$ has $K^{-1}$ in the unit.

Why this happens:

Students mix up the ebullioscopic constant ($K_b$) with the cryoscopic constant ($K_f$) or with the reciprocal relationship in formulas like $\Delta T_b = K_b \cdot m$.

How to avoid:

• Remember the formula:

Here $\Delta T_b$ is in K and $m$ (molality) is in mol/kg.

• Solve for $K_b$:

So units are:

• Key insight: $K_b$ tells you how many Kelvin the boiling point rises per unit molality — so Kelvin is in the numerator, not denominator.

Mistake #2: Reversing the numerator and denominator

What students do wrong:

They pick option (iv) — $K\ mol\ kg^{-1}$ — which has mol in the numerator and kg in the denominator.

Why this happens:

Students misremember the formula as $\Delta T_b = \frac{K_b}{m}$ or confuse molality ($mol/kg$) with its reciprocal ($kg/mol$).

How to avoid:

• Write the formula clearly:

• Molality $m$ has units $mol\ kg^{-1}$.
• For the product to give Kelvin, $K_b$ must cancel the $mol\ kg^{-1}$ part:

• Trick: The unit of $K_b$ is the reciprocal of the unit of molality.

Mistake #3: Picking option (ii) — $mol\ kg\ K^{-1}$

What students do wrong:

They think $K_b$ has mol in the numerator and Kelvin in the denominator.

Why this happens: …