Question

Molal depression constant for a liquid is $2.77^{\circ}\mathrm{C}\ \mathrm{kg}\ \mathrm{mol}^{-1}$, in Kelvin scale its value is

• A. $275.77\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$
• B. $271.77\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$
• C. $2.77\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$
• D. $27.7\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$

Hint:

Always distinguish between an absolute temperature ($T$) and a temperature difference ($\Delta T$). For absolute values, add 273.15 to convert to Kelvin. For intervals, coefficients, or constants (like $K_f$, $K_b$, or specific heat capacities), the numerical value stays exactly the same!
{|c|c|c|} Measurement Type & Celsius Scale & Kelvin Scale Absolute Temperature & $T = 0^{\circ}\mathrm{C}$ & $T = 273.15\ \mathrm{K}$ Temperature Difference & $\Delta T = 1^{\circ}\mathrm{C}$ & $\Delta T = 1\ \mathrm{K}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to convert the given value of the molal depression constant ($K_f = 2.77^{\circ}\mathrm{C}\ \mathrm{kg}\ \mathrm{mol}^{-1}$) from the Celsius scale to the Kelvin scale.

Step 2: Key Formula or Approach:
The molal depression constant describes a temperature change or difference ($\Delta T_f$) per unit molality: $$K_f = \frac{\Delta T_f}{m}$$ The size of one degree Celsius ($1^{\circ}\mathrm{C}$) is exactly identical to the size of one Kelvin ($1\ \mathrm{K}$). Therefore, any temperature difference ($\Delta T$) has the exact same numerical value on both scales: $$\Delta T\ (\text{in }^{\circ}\mathrm{C}) = \Delta T\ (\text{in }\mathrm{K})$$

Step 3: Detailed Explanation:
Because $K_f$ is defined in terms of a temperature interval rather than an absolute value, we do not add 273.15 to the number. A lowering of $2.77^{\circ}\mathrm{C}$ in freezing point is physically identical to a lowering of $2.77\ \mathrm{K}$. Thus, substituting the unit symbols directly while preserving the baseline magnitude yields: $$2.77^{\circ}\mathrm{C}\ \mathrm{kg}\ \mathrm{mol}^{-1} = 2.77\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$$

Step 4: Final Answer:
The value in the Kelvin scale is $2.77\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$, which corresponds to option (C).