The boiling point elevation of a solution is given by the formula:

ΔTb = i * Kb * m

where:
- ΔTb is the change in boiling point,
- i is the van't Hoff factor (which is 2 for NaCl, because it dissociates into 2 ions in solution),
- Kb is the ebullioscopic constant (which is 0.51°C/(mol/kg) for water), and
- m is the molality of the solution (which is the number of moles of solute per kilogram of solvent).

First, we need to calculate the molality of the solution. Since we have 3 mol of NaCl and 1 kg of water, the molality is:

m = 3 mol / 1 kg = 3 mol/kg

Then, we can substitute these values into the formula to find the change in boiling point:

ΔTb = 2 * 0.51°C/(mol/kg) * 3 mol/kg = 3.06°C

The normal boiling point of water is 100°C, so the final boiling point of the water after adding the NaCl would be:

100°C + 3.06°C = 103.06°C

So, the correct answer is C.103.06°C.

Question

The boiling point elevation of a solution is given by the formula:

ΔTb = i * Kb * m

where:
- ΔTb is the change in boiling point,
- i is the van't Hoff factor (which is 2 for NaCl, because it dissociates into 2 ions in solution),
- Kb is the ebullioscopic constant (which is 0.51°C/(mol/kg) for water), and
- m is the molality of the solution (which is the number of moles of solute per kilogram of solvent).

First, we need to calculate the molality of the solution. Since we have 3 mol of NaCl and 1 kg of water, the molality is:

m = 3 mol / 1 kg = 3 mol/kg

Then, we can substitute these values into the formula to find the change in boiling point:

ΔTb = 2 * 0.51°C/(mol/kg) * 3 mol/kg = 3.06°C

The normal boiling point of water is 100°C, so the final boiling point of the water after adding the NaCl would be:

100°C + 3.06°C = 103.06°C

So, the correct answer is C.103.06°C.

Knowee AI · Accepted Answer

The boiling point elevation of a solution is given by the formula:

ΔTb = i * Kb * m

where:
- ΔTb is the change in boiling point,
- i is the van't Hoff factor (which is 2 for NaCl, because it dissociates into 2 ions in solution),
- Kb is the ebullioscopic constant (which is 0.51°C/(mol/kg) for water), and
- m is the molality of the solution (which is the number of moles of solute per kilogram of solvent).

First, we need to calculate the molality of the solution. Since we have 3 mol of NaCl and 1 kg of water, the molality is:

m = 3 mol / 1 kg = 3 mol/kg

Then, we can substitute these values into the formula to find the change in boiling point:

ΔTb = 2 * 0.51°C/(mol/kg) * 3 mol/kg = 3.06°C

The normal boiling point of water is 100°C, so the final boiling point of the water after adding the NaCl would be:

100°C + 3.06°C = 103.06°C

So, the correct answer is C.103.06°C.

What would the final boiling point of water be if 3 mol of NaCl were added to 1 kg of water (Kb = 0.51°C/(mol/kg) for water and i = 2 for NaCl)?A.101.53°CB.98.47°CC.103.06°CD.96.94°C

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