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1 year ago

What is the molarity of a 0.65L solution containing 63 grams of ? The molar mass of NaCl is 58.44 g/mol

Chemistry

1 answer:

ad-work [718]1 year ago

7 0

Explanation:

The molarity of a solution is defined like the number of moles of solute per liters of solution.

molarity = moles of solute/(volume of solution in L)

We know the volume of solution in L.

volume of solution = 0.65 L

To go from the mass of our solute in grams to moles we have to use its molar mass.

mass of NaCl = 63 g

molar mass of NaCl = 58.44 g/mol

moles of NaCl = 63 g * 1 mol/(58.44 g)

moles of NaCl = 1.078 moles

Finally we can find the molarity of the solution

molarity = moles of NaCl/(volume of solution)

molarity = 1.078 moles/(0.65 L)

molarity = 1.66 M

Answer: the molarity of the solution is 1.66 M.

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Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

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What is the mole fraction of NaCl in a mixture containing 7.21 moles NaCl, 9.37 moles KCL, and 3.42

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Answer : The mole fraction of NaCl in a mixture is, 0.360

Explanation : Given,

Moles of $NaCl$ = 7.21 mole

Moles of $KCl$ = 9.37 mole

Moles of $LiCl$ = 3.42 mole

Now we have to calculate the mole fraction of $NaCl$ .

$\text{Mole fraction of }NaCl=\frac{\text{Moles of }NaCl}{\text{Moles of }NaCl+\text{Moles of }KCl+\text{Moles of }LiCl}$

Now put all the given values in this formula, we get:

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