Concentration units (molality): A solution contains 0.30 kmol of a solute dissolved in 600 kg of solvent.\r What is the molality of the solution (in mol per kg of solvent)?

Introduction:
Molality (m) is a temperature-independent concentration unit widely used in thermodynamics and colligative property calculations. It is defined as moles of solute per kilogram of solvent, not per kilogram of solution. Correctly identifying solute amount and solvent mass is crucial.

Given Data / Assumptions:

• Moles of solute n_s = 0.30 kmol.
• Mass of solvent m_solvent = 600 kg.
• Molality m = moles of solute per kilogram of solvent.

Concept / Approach:
Use the direct definition of molality. Units can be kmol/kg or mol/kg; numerically they are consistent as long as moles and kilograms are used. Here we can compute using kmol/kg and interpret the result in mol/kg by noting that 1 kmol/kg equals 1000 mol/kg numerically, but the ratio value is unchanged when reported simply as “molality.”

Step-by-Step Solution:
m = n_s / m_solventSubstitute: m = 0.30 kmol / 600 kgCompute: m = 0.0005 kmol/kg = 0.5 mol/kgTherefore, molality = 0.50

Verification / Alternative check:
A quick proportional check: halving the denominator from 600 to 300 would double molality from 0.5 to 1.0, which is consistent with the formula.

Why Other Options Are Wrong:

• 0.60 or 1 or 2: Do not match the computed ratio.
• 0.05: Off by a factor of 10; a common slip when converting kmol to mol.

Common Pitfalls:
Mistakenly using mass of solution instead of mass of solvent, or converting kmol to mol and then dividing by kilograms again incorrectly.

Final Answer:
0.50