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[PHC - JamCol]

If I make a solution by adding 83 grams of sodium hydroxide to 750 mL of water...
a. What is the molality of sodium hydroxide in this solution?
b. What is the percent by mass of sodium hydroxide in this solution?
c. What is the mole fraction of sodium hydroxide in this solution?

 

[🤖]

a. To calculate the molality of sodium hydroxide in the solution, we need to first calculate the number of moles of sodium hydroxide present in the solution.

Number of moles of NaOH = mass / molar mass
where mass = 83 g and molar mass of NaOH = 40 g/mol + 1 g/mol + 16 g/mol = 40 g/mol (since Na has a mass number of 23 and O has a mass number of 16)

Number of moles of NaOH = 83 g / 40 g/mol = 2.075 mol

Next, we need to calculate the mass of water in the solution, using the density of water at room temperature (which is close to 750 mL).

Mass of water = density x volume
where density of water = 1 g/mL (at room temperature) and volume = 750 mL

Mass of water = 1 g/mL x 750 mL = 750 g

Now, we can calculate the molality of sodium hydroxide in the solution:

Molality = number of moles of solute / mass of solvent (in kg)
where number of moles of solute = 2.075 mol and mass of solvent = 750 g / 1000 g/kg = 0.75 kg

Molality = 2.075 mol / 0.75 kg = 2.767 mol/kg (rounded to three significant figures)

Therefore, the molality of sodium hydroxide in this solution is approximately 2.77 mol/kg.

b. To calculate the percent by mass of sodium hydroxide in the solution, we need to divide the mass of sodium hydroxide by the total mass of the solution and multiply by 100%.

Mass of NaOH = 83 g
Total mass of solution = 83 g + 750 g = 833 g

Percent by mass of NaOH = (mass of NaOH / total mass of solution) x 100%
= (83 g / 833 g) x 100%
= 9.96% (rounded to two significant figures)

Therefore, the percent by mass of sodium hydroxide in this solution is approximately 9.96%.

c. To calculate the mole fraction of sodium hydroxide in the solution, we need to divide the number of moles of sodium hydroxide by the total number of moles of solute and solvent.

Number of moles of solvent (water) = mass / molar mass
= 750 g / 18 g/mol (molar mass of water)
= 41.67 mol

Total number of moles of solute and solvent = 2.075 mol + 41.67 mol = 43.745 mol

Mole fraction of NaOH = number of moles of NaOH / total number of moles
= 2.075 mol / 43.745 mol
= 0.0474 (rounded to four significant figures)

Therefore, the mole fraction of sodium hydroxide in this solution is approximately 0.0474.

 

[Nezhaaaa]

To answer your questions, we need to calculate the molality, percent by mass, and mole fraction of sodium hydroxide in the given solution.

First, let's calculate the molality (m) of sodium hydroxide in the solution:

a. Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is sodium hydroxide (NaOH) and the solvent is water (H2O).

Step 1: Calculate the number of moles of sodium hydroxide (NaOH):We need to convert grams to moles. The molar mass of NaOH is:Na: 22.99 g/molO: 16.00 g/molH: 1.01 g/molTotal molar mass: 22.99 + 16.00 + 1.01 = 40.00 g/mol

Number of moles = mass / molar massNumber of moles of NaOH = 83 g / 40.00 g/mol = 2.075 mol

Step 2: Calculate the mass of water (H2O)ensity of water = 1 g/mLVolume of water = 750 mLMass of water = volume × density = 750 mL × 1 g/mL = 750 g

Step 3: Calculate the molality:Molality (m) = moles of solute / mass of solvent (in kg)Mass of solvent (water) = 750 g = 0.750 kg

Molality (m) = 2.075 mol / 0.750 kg ≈ 2.767 mol/kg

Therefore, the molality of sodium hydroxide in the solution is approximately 2.767 mol/kg.

Now, let's calculate the percent by mass of sodium hydroxide in the solution:

b. Percent by mass is the ratio of the mass of the solute to the mass of the solution, multiplied by 100.

Mass of sodium hydroxide = 83 gMass of the solution = mass of sodium hydroxide + mass of waterMass of the solution = 83 g + 750 g = 833 g

Percent by mass of sodium hydroxide = (mass of sodium hydroxide / mass of the solution) × 100Percent by mass of sodium hydroxide = (83 g / 833 g) × 100 ≈ 9.96%

Therefore, the percent by mass of sodium hydroxide in the solution is approximately 9.96%.

Lastly, let's calculate the mole fraction of sodium hydroxide in the solution:

c. The mole fraction (X) is the ratio of the moles of solute to the total moles of all components in the solution.

Moles of sodium hydroxide = 2.075 molMoles of water = mass of water / molar mass of waterMolar mass of water = 18.015 g/mol

Moles of water = 750 g / 18.015 g/mol ≈ 41.63 mol

Total moles of all components = moles of sodium hydroxide + moles of waterTotal moles of all components = 2.075 mol + 41.63 mol = 43.705 mol

Mole fraction of sodium hydroxide = moles of sodium hydroxide / total moles of all componentsMole fraction of sodium hydroxide = 2.075 mol / 43.705 mol ≈ 0.0475

Therefore, the mole fraction of sodium hydroxide in the solution is approximately 0.0475.