© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Recall the Definitions
⢠Molality (m): Number of moles of solute per 1 kg of solvent.
⢠Molarity (M): Number of moles of solute per 1 L of solution.
Step 2: Identify Given Data
⢠Molality, m = 6.50 \text{ mol kg}^{-1} .
⢠Density of the solution, d = 1.89 \text{ g cm}^{-3} = 1.89 \times 10^3 \text{ g L}^{-1} .
⢠Molar mass of KOH, M_{\text{KOH}} = 39 + 16 + 1 = 56 \text{ g mol}^{-1} .
⢠Desired quantity: Molarity, M (in \text{mol L}^{-1} ).
Step 3: Use the Relationship Between Molality and Molarity
Consider 1 L of solution (volume basis for molarity). This 1 L of solution will have a mass of 1890 g (because the density is 1.89 g mLâ1, i.e., 1.89 Ă 103 g per 1 L). Out of this total mass, a certain amount corresponds to KOH (the solute) and the rest is water (the solvent). We can write the formula relating molality and molarity:
m = \dfrac{1000 \times M}{(1000 \times d) - (M \times M_{\text{solute}})}
Here:
m = 6.5 (molality)
M = molarity (unknown)
d = 1.89 \times 10^3 \, \text{g L}^{-1}
M_{\text{solute}} = 56 \, \text{g mol}^{-1} (for KOH)
Step 4: Plug In the Numbers and Solve
Substitute the values in the formula:
6.5 = \dfrac{1000 \times M}{1890 - 56M}
Cross-multiply and simplify:
6.5 \times (1890 - 56M) = 1000M
12285 - 364M = 1000M
12285 = 1000M + 364M = 1364M
M = \dfrac{12285}{1364} \approx 9
Step 5: State the Final Answer
Therefore, the molarity of the solution is approximately 9 mol Lâ1.
