The boiling point of water at 735mmHg is 99.073°C. What mass of BaCl2 should be added to 2.28kg water to increase the boiling point to 100.000°C ?

Kb for water = 0.510 K kg mol-1
2589 g
777 g
734 g
288 g
863 g

delta T = i*Kb*m

Solve for m

molality = moles/kg
Solve for moles

moles = grams/molar mass
Solve for grams.

dT= 100- 99.073= 0.927

dT= m Kb i
0.972= m*0.51*3= 0.972/0.510*3
=0.606 mol BaCl2/ 2.28kg
0.606*2.28kg= 1.38g
mass BaCl2= 1.38g*208.23g/mol
=288g

To solve this problem, we can use the equation for calculating the boiling point elevation:

∆T = Kb * m

Where:
∆T = change in boiling point
Kb = boiling point elevation constant (0.510 K kg mol-1 for water)
m = molality of the solution (moles of solute per kilogram of solvent)

First, we need to calculate the molality of the solution:

molality (m) = moles of solute / kg of solvent

To find the moles of BaCl2, we need to use its molar mass. The molar mass of BaCl2 is:
Ba = 137.33 g/mol
Cl = 35.453 g/mol (x 2) = 70.906 g/mol
Total molar mass = 137.33 + 70.906 = 208.236 g/mol

Now we can calculate the moles of BaCl2 needed:
moles of BaCl2 = mass of BaCl2 / molar mass of BaCl2

mass of BaCl2 = molality * kg of water

Since we know the change in boiling point (∆T) and the molality of the solution (m), we can rearrange the equation to solve for the mass of BaCl2:

mass of BaCl2 = ∆T * m / Kb * kg of water

Given:
∆T = 100.000°C - 99.073°C = 0.927°C
Kb = 0.510 K kg mol-1
kg of water = 2.28 kg

Plugging in the values:

mass of BaCl2 = 0.927°C * (molality * kg of water) / (0.510 K kg mol-1)

Now, let's calculate the mass of BaCl2 needed:

molality = moles of BaCl2 / kg of water
molality = (mass of BaCl2 / molar mass of BaCl2) / kg of water

mass of BaCl2 = ∆T * ((mass of BaCl2 / molar mass of BaCl2) / kg of water) / Kb

mass of BaCl2 * (molar mass of BaCl2) = ∆T * ((mass of BaCl2) / kg of water) / Kb

mass of BaCl2 = (∆T * (kg of water) * molar mass of BaCl2) / (Kb - ∆T)

mass of BaCl2 = (0.927°C * 2.28 kg * 208.236 g/mol) / (0.510 K kg mol-1 - 0.927°C)

Calculating this gives us:

mass of BaCl2 = 863.93 g (approximately)

Therefore, the correct answer is 863 g.

To find the mass of BaCl2 needed to increase the boiling point of water, we can use the formula:

∆Tb = Kb * m * i

Where:
∆Tb = change in boiling point
Kb = boiling point elevation constant for water (0.510 K kg mol-1)
m = molality of the solution (moles of solute per kilogram of solvent)
i = van't Hoff factor (number of particles into which the solute dissociates)

We need to find the molality of the solution. Molality (m) is defined as the number of moles of solute divided by the mass of the solvent in kilograms.

Given:
Boiling point of water at atmospheric pressure = 100.000°C
Boiling point of water at 735 mmHg = 99.073°C
Change in boiling point (∆Tb) = 100.000°C - 99.073°C = 0.927°C

We can now apply the formula to calculate the molality (m):

0.927 = 0.510 * m * i

We need to find the mass of BaCl2, so we need to find the moles of BaCl2 first.

Molar mass of BaCl2 = molar mass of Ba + 2 * molar mass of Cl = 137.33 + 2 * 35.45 = 208.23 g/mol

Number of moles of BaCl2 = mass / molar mass = X g / 208.23 g/mol

Now, we can calculate the molality (m) using the given equation:

0.927 = 0.510 * (X g / 208.23 g/mol) * i

Simplifying the equation, we find:

X g = (0.927 * 208.23 g/mol) / (0.510 * i)

Since the van't Hoff factor (i) of BaCl2 is 3 (as it dissociates into 3 ions: Ba2+ and 2 Cl-), we can substitute the value of i into the equation:

X g = (0.927 * 208.23 g/mol) / (0.510 * 3)

Solving this equation will give us the mass of BaCl2 required to increase the boiling point to 100.000°C.

Note: We need the units of molality (m) and mass to be consistent, so make sure to convert the mass of water to kilograms before proceeding with the calculations.

Let's calculate the mass of BaCl2:

X g = (0.927 * 208.23 g/mol) / (0.510 * 3)
X g = 287.78 g

Therefore, the mass of BaCl2 that should be added to 2.28 kg of water to increase the boiling point to 100.000°C is approximately 288 g.

So, the correct answer is 288 g.