Question

a chemist combined chloroform (chcl₃) and acetone (c₃h₆o) to create a solution where the mole fraction of chloroform, $chi_{chloroform}$, is 0.203. the densities of chloroform and acetone are 1.48 g/ml and 0.791 g/ml, respectively.calculate the molarity of the solution. assume the volumes are additive.molarity: ____ mincorrect answercalculate the molality of the solution.molality: 4.38 ____ mcorrect answer

Explanation:

Step1: Define mole fractions

Let $n_{\text{CHCl}_3} = 0.203\ \text{mol}$, so $n_{\text{C}_3\text{H}_6\text{O}} = 1 - 0.203 = 0.797\ \text{mol}$

Step2: Calculate molar masses

Molar mass of $\text{CHCl}_3$: $12.01 + 1.008 + 3\times35.45 = 119.378\ \text{g/mol}$
Molar mass of $\text{C}_3\text{H}_6\text{O}$: $3\times12.01 + 6\times1.008 + 16.00 = 58.078\ \text{g/mol}$

Step3: Find masses of each component

Mass of $\text{CHCl}_3$: $m_{\text{CHCl}_3} = 0.203\ \text{mol} \times 119.378\ \text{g/mol} = 24.2337\ \text{g}$
Mass of $\text{C}_3\text{H}_6\text{O}$: $m_{\text{C}_3\text{H}_6\text{O}} = 0.797\ \text{mol} \times 58.078\ \text{g/mol} = 46.2882\ \text{g}$

Step4: Calculate volumes of each component

Volume of $\text{CHCl}_3$: $V_{\text{CHCl}_3} = \frac{24.2337\ \text{g}}{1.48\ \text{g/mL}} = 16.3741\ \text{mL}$
Volume of $\text{C}_3\text{H}_6\text{O}$: $V_{\text{C}_3\text{H}_6\text{O}} = \frac{46.2882\ \text{g}}{0.791\ \text{g/mL}} = 58.5186\ \text{mL}$

Step5: Find total solution volume

$V_{\text{total}} = 16.3741 + 58.5186 = 74.8927\ \text{mL} = 0.0748927\ \text{L}$

Step6: Compute molarity

Molarity $= \frac{n_{\text{CHCl}_3}}{V_{\text{total}}} = \frac{0.203\ \text{mol}}{0.0748927\ \text{L}}$

Answer:

2.71 M