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13 days ago

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A 7.27-gram sample of a compound is dissolved in 250 grams of benzene. The freezing point of this solution is 1.02°C below that

of pure benzene. What is the molar mass of this compound? (Note: Kf for benzene = 5.12°C/m.) Ignore significant figures for this problem:_________.
a. 36.5 g/mol
b. 146 g/mol
c. 292 g/mol
d. 5.79 g/mol
e. 73.0 g/mol

1 answer:

lorasvet [2.6K]13 days ago

5 0

Answer:146 g/mol → option b.

Explanation:

This problem deals with freezing point depression. The colligative property formula for this is:

ΔT = Kf. m. i

We assume i = 1, indicating our compound is non-electrolytic.

ΔT represents the difference in freezing temperature of the pure solvent and the solution = 1.02 °C

m stands for molality (mol of solute per kg of solvent)

We convert the mass of the solvent (benzene) to kg → 250 g. 1 kg/1000 = 0.250 kg.

Substituting gives us → 1.02°C = 5.12°C/mol/kg. mol/ 0.250kg. 1

1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg

0.19922 mol/kg = mol/ 0.250kg

mol = 0.19922. 0.250kg → 0.0498 mol

Calculating molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol

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A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2624]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

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11 days ago

Which best describes the relationships between subatomic particles in any neutral atom? * 1 point the number of electrons equals

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The atomic number corresponds to the number of protons

Protons are denoted as P and Electrons as E P = E

The atomic mass equals the sum of Neutrons and Protons

Atomic number = atomic mass = neutrons

P = E

Atomic mass - atomic number = Neutrons

Example:

Calcium consists of 20 Protons 20P = 20E

Atomic mass - atomic number = neutron count:)

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11 days ago

How many moles of nitrogen gas are there in 6.8 liters at room temperature and pressure (293 K and 100 kPa)?

Alekssandra [2891]

Utilize the ideal gas law:

n = PV / RT

P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
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Substituting all values leads to:
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1 month ago

A substance has a density of 4.5 g/cm3. The substance is heated and its density recalculated. What is the newly recorded density

lions [2782]

Response:

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Clarification:

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9 days ago

What are the respective central-metal oxidation state, coordination number, and overall charge on the complex ion in Na2[Cr(NH3)

lions [2782]

Answer:

Central metal oxidation state: +2

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Overall charge: -2

Explanation:

For the ion complex:

Na₂[Cr(NH₃)₂(NCS)₄]

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NH₃ acts as a neutral ligand, while NCS carries a negative charge.

The entire complex has a charge of:

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Since each NCS contributes -1 and there are four NCS, the Cr must possess an oxidation state of +2 to achieve an overall charge of -2.

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I trust this clarifies the matter!

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1 month ago