When 2.25 g of sodium hydroxide (NaOH) was dissolved in 150.00 g of water a value of 11.00oC was obtained for ΔT.1. Calculate the molarity of the sodium hydroxide solution.
2. Calculate the value (calories) for the heat of solution of 2.25 g of NaOH.
3. Calculate the number of calories that would be produced if one mole of sodium hydroxide was dissolved. (ΔHsolnNaOH)

Answers

Answer 1
Answer:

Answer:

For 1: The molarity of sodium hydroxide solution is 0.375 M

For 2: The amount of heat absorbed by solution is 1674.75 Cal

For 3: The enthalpy change of the reaction when 1 mole of NaOH is dissolved is 1674.75 Cal

Explanation:

  • For 1:

To calculate mass of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Mass of water = 150.00 g

Putting values in above equation, we get:

1g/mL=\frac{150.00g}{\text{Volume of water}}\n\n\text{Volume of water}=(1g/mL* 150.00g)=150.00mL

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}

We are given:

Mass of solute (sodium hydroxide) = 2.25 g

Molar mass of sodium hydroxide = 40 g/mol

Volume of solution = 150.00 mL

Putting values in above equation, we get:

\text{Molarity of solution}=(2.25g* 1000)/(40g/mol* 150mL)\n\n\text{Molarity of solution}=0.375M

Hence, the molarity of sodium hydroxide solution is 0.375 M

  • For 2:

To calculate the amount of heat absorbed, we use the equation:

q=mc\Delta T

where,

m = mass of solution = (2.25 + 150) = 152.25 g

c = specific heat capacity of water = 1 Cal/g.°C

\Delta T = change in temperature = 11°C

Putting values in above equation, we get:

q=152.25g* 1Cal/g.^oC* 11^oC\n\nq=1674.75Cal

Hence, the amount of heat absorbed by solution is 1674.75 Cal

  • For 3:

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_(rxn)=(q)/(n)

where,

q = amount of heat absorbed = 16

74.75 Cal

n = number of moles = 1 mole

\Delta H_(rxn) = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_(rxn)=(1674.75Cal)/(1mol)=1674.75Cal/mol

Hence, the enthalpy change of the reaction when 1 mole of NaOH is dissolved is 1674.75 Cal

Answer 2
Answer:

Final answer:

The molarity of the sodium hydroxide is 0.375 M. The heat of solution of the sodium hydroxide is -1650 cal, and the heat of solution per mole of sodium hydroxide is -29333.33 cal/mol.

Explanation:

To answer your questions, we first need to convert the mass of the sodium hydroxide (NaOH) to moles. Sodium hydroxide has a molar mass of approximately 40 g/mol, so 2.25 g is 0.05625 mol.

1. The molarity of the solution is the number of moles of solute per liter of solution. Given that the solution was made up in 150.00 g of water, which is approximately 0.150 L (since the density of water is approximately 1 g/mL), the molarity is 0.05625 mol / 0.150 L = 0.375 M.

2. The heat of solution can be calculated using the equation q = m * c * ΔT, where m is the mass of the water, c is the specific heat capacity of the water (approximately 1 cal/g°C), and ΔT is the change in temperature. Plugging in the known values, q = 150.00 g * 1 cal/g°C * 11°C = 1650 cal. This is the heat absorbed by the water and so the heat of solution of NaOH is -1650 cal (as the process of dissolving is exothermic).

3. The heat of solution per mole of sodium hydroxide can be calculated by dividing the total heat of solution by the number of moles of sodium hydroxide. So ΔHsoln NaOH = -1650 cal / 0.05625 mol = -29333.33 cal/mol.

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Answers

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1.01 x 10^-3 moles

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Answers

Answer:

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Explanation:

Step 1: Given and required data

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We will use the following expression.

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Answers

Answer:

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Proper equation of the reaction:

                    αC₆H₁₄ + βO₂ → γCO₂ + δH₂O

This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:

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γ = 6

δ = (n)/(2) = (14)/(2) = 7

The complete reaction equation is therefore given as:

                   C₆H₁₄ + 9(1)/(2)O₂ → 6CO₂ + 7H₂O

To express as whole number integers, we multiply the coefficients through by 2:

                  2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

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