Question

If 6 g of solute dissolved in 100 g of water lowers the freezing point by 0.93 K. What is molar mass of solute? ($\mathrm{K}_f = 1.86\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$)

• A. $120\ \mathrm{g}\ \mathrm{mol}^{-1}$
• B. $60\ \mathrm{g}\ \mathrm{mol}^{-1}$
• C. $90\ \mathrm{g}\ \mathrm{mol}^{-1}$
• D. $180\ \mathrm{g}\ \mathrm{mol}^{-1}$

Hint:

Look for clean numerical factors before multiplying everything out! Notice that $1.86$ is exactly twice the value of $0.93$. Recognizing this relationship allows you to simplify the fraction to a clean factor of $2$ instantly, making the rest of the math effortless.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the mass of a solute ($W_2 = 6\ \mathrm{g}$) dissolved in a mass of water solvent ($W_1 = 100\ \mathrm{g}$). This solution lowers the freezing point by $\Delta T_f = 0.93\ \mathrm{K}$. Given the cryoscopic constant ($\mathrm{K}_f = 1.86\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1}$), we need to compute the solute's molar mass ($M_2$).

Step 2: Key Formula or Approach:
The relationship for the lowering of the freezing point is defined by: $$\Delta T_f = \mathrm{K}_f \cdot m$$ Where molality ($m$) is calculated as: $$m = \frac{W_2 \times 1000}{M_2 \times W_1}$$ Combining these into a single equation to isolate and solve for the molar mass ($M_2$) yields: $$M_2 = \frac{\mathrm{K}_f \times W_2 \times 1000}{\Delta T_f \times W_1}$$

Step 3: Detailed Explanation:
Substitute the given values directly into the rearranged equation: $$M_2 = \frac{1.86 \times 6 \times 1000}{0.93 \times 100}$$ Simplify the fractional parts to make the arithmetic straightforward: $$\frac{1000}{100} = 10$$ $$\frac{1.86}{0.93} = 2$$ Now, multiply the remaining simplified values: $$M_2 = 2 \times 6 \times 10 = 120\ \mathrm{g}\ \mathrm{mol}^{-1}$$

Step 4: Final Answer:
The molar mass of the solute is $120\ \mathrm{g}\ \mathrm{mol}^{-1}$, which matches option (A).