Chemistry, 30.07.2019 00:30 queenz58

Acompound weighing 0.458 g is dissolved in 30.0 g of acetic acid. the freezing point of the solution is found to be 1.50 k below that of the pure solvent. calculate the molar mass of the compound.

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Answer from: grumpygreengori

Molar mass of compound = 38.17 g/mol

Explanation:

The mass of compound dissolved = 0.458 g

The mass of acetic acid taken = 30.0g = 0.03 kg

the depression in freezing point =1.50 K or 1.50 ⁰C

the relation between depression in freezing point and molality is:

Depression in freezing point = Kf X molality

Where Kf= cryoscopic constant = 3.90 ⁰C Kg/mol

Putting values

1.50 = 3.90 X molality

$molality=\frac{1.50}{3.90}=0.385$

molality is moles of solute per Kg of solvent

$molality=\frac{moles}{massofsolvent}=\frac{moles}{0.03}=0.385$

moles = 0.385 X 0.03 = 0.012

$moles=\frac{mass}{molarmass}$

$molarmass=\frac{mass}{moles}=\frac{0.458}{0.012}= 38.17g/mol$

Answer from: brianna8739

The molar mass of the compound is 39.69 g/mol

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$\Delta T_f$ = Depression in freezing point = 1.50 K = 1.50°C (Change remains constant)

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 3.90°C/m

$m_{solute}$ = Given mass of solute = 0.458 g

$M_{solute}$ = Molar mass of solute (glucose) = ? g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 30.0 g

Putting values in above equation, we get:

$1.50^oC=1\times 3.90^oC/m\times \frac{0.458\times 1000}{\text{Molar mass of solute}\times 30.0}\\\\\text{Molar mass of solute}=\frac{1\times 3.90\times 0.458\times 1000}{1.50\times 30.0}=39.69g/mol$

Hence, the molar mass of the compound is 39.69 g/mol

Answer from: Quest

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.

Answer from: Quest

Qand k are only the same when the forward rate of the reaction has reached zero.