Chemistry - Concentration Terms Question with Solution | TestHub

To find molality, we need moles of solute and mass of solvent.

Given:

Molarity ($M$) = $2.05\mathrm{mol}{\mathrm{L}}^{-1}$

Density of solution ($\rho$) = $1.02\mathrm{g}{\mathrm{mL}}^{-1}$

Solute: Acetic acid (CH₃COOH)

Molar mass of CH₃COOH = $12\times 2+1\times 4+16\times 2=60\mathrm{g}{\mathrm{mol}}^{-1}$

Step 1: Assume a volume of solution.

Let's assume $1\mathrm{L}$ (or $1000\mathrm{mL}$) of solution.

Step 2: Calculate moles of solute.

Moles of CH₃COOH = Molarity $\times$ Volume of solution

Moles of CH₃COOH = $2.05\mathrm{mol}{\mathrm{L}}^{-1}\times 1\mathrm{L}=2.05\mathrm{mol}$

Step 3: Calculate mass of solute.

Mass of CH₃COOH = Moles $\times$ Molar mass

Mass of CH₃COOH = $2.05\mathrm{mol}\times 60\mathrm{g}{\mathrm{mol}}^{-1}=123\mathrm{g}$

Step 4: Calculate mass of solution.

Mass of solution = Density $\times$ Volume of solution

Mass of solution = $1.02\mathrm{g}{\mathrm{mL}}^{-1}\times 1000\mathrm{mL}=1020\mathrm{g}$

Step 5: Calculate mass of solvent (water).

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent = $1020\mathrm{g}-123\mathrm{g}=897\mathrm{g}$

Convert mass of solvent to kg: $897\mathrm{g}=0.897\mathrm{kg}$

Step 6: Calculate molality ($m$).

Molality = $\frac{Molesofsolute}{Massofsolvent(inkg)}$

Molality = $\frac{2.05\mathrm{mol}}{0.897\mathrm{kg}}\approx 2.285\mathrm{mol}{\mathrm{kg}}^{-1}$

Therefore, the molality of the solution is approximately $2.28\mathrm{mol}{\mathrm{kg}}^{-1}$.

The final answer is $\boxed{2.28\mathrm{mol}{\mathrm{kg}}^{-1}}$.