Question

An aqueous solution containing $0.250\mathrm{mol}\mathrm{Q}$, a strong electrolyte, in $5.00\times {10}^2\mathrm{g}$ water freezes at $-{2.79}^{\circ }\mathrm{C}$. What is the van't Hoff factor for Q? The molal freezing-point depression constant for water is ${1.86}^{\circ }\mathrm{C}\cdot \mathrm{kg}/\mathrm{mol}$. What is the formula of $\mathrm{Q}$ if it is $38.68\mathrm{\%}$ chlorine by mass and there are twice as many anions as cations in one formula unit of $\mathrm{Q}$ ?

Short answer

The van't Hoff factor for Q is approximately 3. The formula of Q is QCl2, with Q having a molar mass of 43.89 g/mol.

Step-by-step solution

Calculate molality of the solution

The formula for molality is: $molality=\frac{molesofsolute}{massofsolventinkg}$ Since we have 0.250 moles of Q in 500 g or 0.500 kg of water, molality is: $molality=\frac{0.250mol}{0.500kg}=0.50mol/kg$

Calculate the experimental freezing-point depression

The freezing-point depression equation is: $\Delta T_f=K_f\times molality\times i$ Where: ΔTf = freezing-point depression, Kf = molal freezing-point depression constant (1.86 °C·kg/mol), molality = calculated in step 1 (0.50 mol/kg), i = van't Hoff factor (unknown for now) We have the experimental freezing-point depression: -2.79 °C.

Find the van't Hoff factor

Rearrange the freezing-point depression equation to solve for i: $i=\frac{\Delta T_f}{K_f\times molality}$ Now substitute the values: $i=\frac{-2.79°C}{1.86°C·kg/mol\times 0.50mol/kg}=2.99\approx 3$ The van't Hoff factor is approximately 3.

Determine the empirical formula of Q

Since there are twice as many anions as cations and the van't Hoff factor is 3, we can say that the empirical formula of Q is QX2, where X is the anion.

Find the formula of Q

Given that Q is 38.68% chlorine by mass and the empirical formula is QX2, we can find the molar masses of Q and X: Let the molar mass of Q be M1 and the molar mass of X be M2. Since QX2 is 38.68% chlorine, we can set up the following equation: $\frac{M1}{M1+2\cdot M2}=0.3868$ Since chlorine is the anion and there are two of them, M2 is the molar mass of chlorine (35.45 g/mol): $\frac{M1}{M1+2\cdot 35.45g/mol}=0.3868$ Now solve for M1: $M1=43.89g/mol$ The formula of Q is therefore QCl2, with Q having a molar mass of 43.89 g/mol.

Key concepts

Molality Calculation

Understanding molality is crucial when working with solutions in chemistry. It's a measure of the concentration of a solute in a solvent, defined as the number of moles of solute per kilogram of solvent. To calculate molality, use the formula:
$$molality=\frac{molesofsolute}{massofsolventinkg}$$
For example, in the given exercise, a solution contains 0.250 moles of a solute Q in 0.500 kg of water, which gives us a molality of 0.50 molal. This value helps us to further determine how the solute affects the solvent's properties, such as the freezing point.

Freezing-point Depression

The phenomenon of freezing-point depression refers to the lowering of the freezing temperature of a pure solvent when a solute is dissolved in it. This is a colligative property, meaning it depends on the number particles the solute contributes to the solution, expressed by the van't Hoff factor (i). The formula to compute the depression is:
$$\mathrm{\Delta }{T}_f=K_f\times molality\times i$$

Calculating Experimental Freezing-Point Depression

Take the change in freezing temperature observed, the molal freezing-point depression constant for the solvent (water, in this case, which is 1.86 °C·kg/mol), and the molality (previously calculated). By knowing the freezing-point depression, you can solve for the van't Hoff factor for the solute, as done in the exercise, indicating the number of particles into which the solute dissociates in solution.

Empirical Formula Determination

The empirical formula represents the simplest whole-number ratio of the elements in a compound. To determine an empirical formula, it's essential to know the proportion of each element by mass and the molar masses of the elements. In our exercise, the van't Hoff factor suggested Q dissociates into three ions, indicating a compound of QX2, assuming X to be the anion. This kind of deduction is fundamental in identifying unknown substances and understanding their chemical behavior.

Percent Composition by Mass

The percent composition by mass of each element within a compound is a critical concept, showing the mass percentage of each constituent element. The calculation often involves dividing the mass of a particular element by the total molar mass of the compound and multiplying by 100%. For instance, the formula QX2 is said to be 38.68% chlorine by mass. Such information is integral for further calculations, like those used to deduce the molar mass of element Q in our exercise, leading to our understanding of the compound's formula.