Question:
A solution is prepared by adding 124 g of ethylene glycol (molar mass = 62 g mol\(^{-1}\)) to \( x \) g of water to get a 10 m solution. What is the value of \( x \) (in g)?
A solution is prepared by adding 124 g of ethylene glycol (molar mass = 62 g mol\(^{-1}\)) to \( x \) g of water to get a 10 m solution. What is the value of \( x \) (in g)?
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Molality (m) is defined as moles of solute per kg of solvent: \( m = \frac{n_{\text{solute}}}{\text{mass}_{\text{solvent (kg)}}} \). Be careful to convert grams to kilograms in denominator!
Updated On: Jun 6, 2025
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The Correct Option is D
Solution and Explanation
Step 1: Understand the given data
- Mass of solute (ethylene glycol) = 124 g
- Molar mass of ethylene glycol = 62 g/mol
- Molality (\( m \)) = 10 m
- Mass of solvent (water) = \( x \) g
Step 2: Use molality formula
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \Rightarrow 10 = \frac{124/62}{x/1000} \]
Step 3: Solve for \( x \)
\[ 10 = \frac{2}{x/1000} \Rightarrow 10 = \frac{2000}{x} \Rightarrow x = 200 \text{ g} \]
Final Answer:
200 g
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