Question

The solution containing 6 g urea (molar mass 60) per $\text{dm}^3$ of water and another solution containing 9 g of solute A per $\text{dm}^3$ water freezes at same temperature. What is molar mass of A?

• A. 90
• B. 180
• C. 54
• D. 120

Hint:

When two solutions with the same solvent share a colligative property value, you can equate their mole fractions directly using a simple ratio: $\frac{W_1}{M_1} = \frac{W_2}{M_2}$. Plunking numbers in gives $\frac{6}{60} = \frac{9}{M_2}$, which immediately yields $M_2 = 90$ with minimal calculation!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two different aqueous solutions: one containing urea and the other containing an unknown solute A. Since both solutions freeze at the exact same temperature, they experience the same depression in freezing point ($\Delta T_f$). We need to determine the molar mass of solute A.

Step 2: Key Formula or Approach:
The depression in freezing point is a colligative property given by the formula:
$$\Delta T_f = K_f \times m$$ Where $K_f$ is the cryoscopic constant of the solvent and $m$ is the molality. For highly dilute solutions, if the volume of the solvent is identical ($1\ \text{dm}^3$), the freezing point depression will match when the total number of moles of solute dissolved per unit volume is identical (equal molar concentrations):
$$\Delta T_{f(\text{urea})} = \Delta T_{f(\text{A})} \implies n_{\text{urea}} = n_{\text{A}}$$

Step 3: Detailed Explanation:
Let's find the number of moles of urea dissolved in $1\ \text{dm}^3$:
Mass of urea ($W_{\text{urea}}$) = $6\ \text{g}$
Molar mass of urea ($M_{\text{urea}}$) = $60\ \text{g\ mol}^{-1}$
$$n_{\text{urea}} = \frac{W_{\text{urea}}}{M_{\text{urea}}} = \frac{6}{60} = 0.1\ \text{mol}$$ Since both solutions are prepared in exactly $1\ \text{dm}^3$ of water and freeze at the same point, the number of moles of solute A must be equal to the number of moles of urea:
$$n_{\text{A}} = n_{\text{urea}} = 0.1\ \text{mol}$$ We know that the mass of solute A ($W_{\text{A}}$) is $9\ \text{g}$. Let's use the mole formula to solve for its molar mass ($M_{\text{A}}$):
$$n_{\text{A}} = \frac{W_{\text{A}}}{M_{\text{A}}} \implies 0.1 = \frac{9}{M_{\text{A}}}$$ $$M_{\text{A}} = \frac{9}{0.1} = 90\ \text{g\ mol}^{-1}$$ This matches option (A).

Step 4: Final Answer:
The molar mass of solute A is 90, which corresponds to option (A).