A solution of nonvolatile solute is obtained by dissolving 15 g in 200 mL water has depression in freezing point 0.75 K. Calculate the molar mass of solute if cryoscopic constant of water is 1.86 K kg mol$^{-1}$.
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Remember the formula for depression in freezing point, $\Delta T_{f} = K_{f} \cdot m$. Ensure consistency in units, especially converting mass of solvent to kilograms or using the factor of 1000 if solvent mass is in grams.
Step 1: Identify the given values and the formula for depression in freezing point.\ Given:\ Mass of solute ($W_{2}$) = $15 \text{ g}$\ Volume of water = $200 \text{ mL}$. Assuming density of water is $1 \text{ g/mL}$, mass of water ($W_{1}$) = $200 \text{ g} = 0.200 \text{ kg}$.\ Depression in freezing point ($\Delta T_{f}$) = $0.75 \text{ K}$\ Cryoscopic constant of water ($K_{f}$) = $1.86 \text{ K kg mol}^{-1}$\ We need to calculate the molar mass of solute ($M_{2}$).\ The formula for depression in freezing point is:\ \[\Delta T_{f} = K_{f} \cdot m\]\ Where $m$ is the molality of the solution. Molality is defined as moles of solute per kilogram of solvent.\ \[m = \frac{\text{moles of solute{\text{mass of solvent (kg) = \frac{W_{2}/M_{2{W_{1} \text{ (in kg)\]\ Step 2: Combine the formulas and rearrange to solve for molar mass ($M_{2}$).\ Substituting the molality expression into the $\Delta T_{f}$ equation:\ \[\Delta T_{f} = K_{f} \cdot \frac{W_{2}/M_{2{W_{1\]\ Rearranging for $M_{2}$:\ \[M_{2} = \frac{K_{f} \cdot W_{2{\Delta T_{f} \cdot W_{1\]\ If $W_{1}$ is in grams, the formula is often written as:\ \[M_{2} = \frac{1000 \cdot K_{f} \cdot W_{2{\Delta T_{f} \cdot W_{1} \text{ (in g)\]\ Step 3: Substitute the values and calculate $M_{2$.}\ Using the formula with $W_{1}$ in grams:\ \[M_{2} = \frac{1000 \cdot 1.86 \text{ K kg mol}^{-1} \cdot 15 \text{ g{0.75 \text{ K} \cdot 200 \text{ g\]\ \[M_{2} = \frac{27900}{150}\]\ \[M_{2} = 186 \text{ g mol}^{-1}\]\ Step 4: Compare with the given options.\ The calculated molar mass is $186 \text{ g mol}^{-1}$, which matches Option C.