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Ne4ueva [31]
3 years ago
5

disolvemos 25 g de hidroxido de sodio en agua hasta obtener 500 cm3 de disolucion. ¿Cual es la concentracion molar de la disoluc

ion?
Chemistry
2 answers:
WARRIOR [948]3 years ago
4 0

Answer:

1.3 M

Explanation:

<em>We dissolve 25 g of sodium hydroxide en water until 500 cm³ of solution are reached. What is the molar concentration of the solution?</em>

<em />

Step 1: Calculate the moles of sodium hydroxide (solute)

The molar mass of sodium hydroxide is 40.00 g/mol.

25 g \times \frac{1mol}{40.00g} =0.63mol

Step 2: Calculate the liters of solution

We will use the following relations:

  • 1 cm³ = 1 mL
  • 1 L = 1,000 mL

500cm^{3} \times \frac{1mL}{1cm^{3} } \times \frac{1L}{1,000mL} =0.500L

Step 3: Calculate the molarity of the solution

M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.63mol}{0.500L} =1.3 M

The molarity of the solution is 1.3 M.

LiRa [457]3 years ago
3 0

Answer:

M=1.25M

Explanation:

Hola,

Dado que la molaridad de una solución es definida en términos de las moles de soluto y el volumen de la solución en litros:

M=\frac{n}{V}

Primero debemos calcular las moles de hidróxido de sodio, teniendo en cuenta que su masa molar es de 40 g/mol:

n=25gNaOH*\frac{1molNaOH}{40gNaOH}=0.625mol

Luego, el volumen el litros:

V=500cm^3*\frac{1L}{1000cm^3} =0.500L

Así, calculamos la concentración molar de la solución:

M=\frac{0.625mol}{0.500L}\\ \\M=1.25M

Saludos!

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1) List the known and unknown quantities.

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

<em>2.1- List the known and unknown quantities.</em>

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

<em>2.2- Convert grams of O2 to moles of O2.</em>

The molar mass of O2 is 31.9988 g/mol.

mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2

<em>2.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>2.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}P_{O_2}=1.57\text{ }atm

<em>The pressure of O2 is 1.57 atm.</em>

3) Find the pressure of F2.

<em>3.1- List the known and unknown quantities.</em>

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- <em>Convert grams of F2 to moles of F2.</em>

The mmolar mass of F2 is 37.9968 g/mol.

mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2

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Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>3.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}P_{F_2}=1.13\text{ }atm

<em>The pressure of F2 is 1.13 atm.</em>

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

<em>4.1- Set the equation.</em>

P_T=P_A+P_B

4.2- Plug in the known quantities.

P_T=1.57\text{ }atm+1.13\text{ }atmP_T=2.7\text{ }atm

<em>The total pressure in the container is </em>2.7 atm<em>.</em>

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Explanation:

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