Chemistry, 12.03.2020 17:25, aubreymoore4553

What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressure depression of 0.734 mmHg at 298 ∘C? (The vapor pressure of water at 298 K is 23.76 mmHg.)

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answered: jrsavala559p9969g

The Van't Hoff factor for KCl is 1.74

Explanation:

We are given:

Molality of solution = 1 m

This means that 1 mole of a solute is present in 1 kg of solvent (water) or 1000 grams of water

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of solute = 1 moles

Total moles = [1 + 55.56] = 56.56 moles

Putting values in above equation, we get:

$\chi_{(solute)}=\frac{1}{56.56}=0.0177$

The equation used to calculate relative lowering of vapor pressure follows:

$\frac{p^o-p_s}{p^o}=i\times \chi_{solute}$

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure = 0.734 mmHg

i = Van't Hoff factor = ?

$\chi_{solute}$ = mole fraction of solute = 0.0177

p^o = vapor pressure of pure water = 23.76 torr

Putting values in above equation, we get:

$\frac{0.734}{23.76}=i\times 0.0177\\\\i=1.74$

Hence, the Van't Hoff factor for KCl is 1.74

answered: Guest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

answered: Guest

Ineed a picture or something

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What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressur...