Question:

What is the molality of a solution containing 300 mg of urea (molar mass 60) dissolved in 30 g of water?

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Molality is calculated by dividing the number of moles of solute by the mass of the solvent in kilograms.
Updated On: Jan 26, 2026
  • 0.133 m
  • 0.825 m
  • 0.498 m
  • 0.166 m
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The Correct Option is D

Solution and Explanation

Step 1: Understanding molality.
Molality is defined as the number of moles of solute per kilogram of solvent. The formula for molality \( m \) is: \[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} \] Step 2: Calculating moles of urea.
The molar mass of urea is given as 60 g/mol. First, convert the mass of urea from mg to g: \[ 300 \, \text{mg} = 0.300 \, \text{g} \] Now, calculate the number of moles of urea: \[ \text{moles of urea} = \frac{0.300}{60} = 0.005 \, \text{mol} \] Step 3: Calculating molality.
The mass of water is 30 g, which is equivalent to 0.030 kg. Now, calculate the molality: \[ m = \frac{0.005}{0.030} = 0.166 \, \text{mol/kg} \] Step 4: Conclusion.
The correct answer is (D) 0.166 m.
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