# calculate the boiling point elevation of a solution containing 2.94 g sodium chloride and 114 g water.?

Please show work.

### 4 Answers

- pisgahchemistLv 79 months ago
Boiling point elevation ....

ΔT = (i)(Kb)(m)

ΔT = (2)(0.512 C/m)(2.94g NaCl x (1 mol NaCl / 58.5g NaCl) / 0.114 kg)

ΔT = 0.451 C

For 2.94 g sodium chloride dissolved in 114 g of water (molality = 0.441m), the boiling point elevation will be 0.451 C degrees.

Should you actually do this, you would find that the boiling point elevation is slightly less than that which is predicted. That's because the van't Hoff factor (i) is not quite "2" for NaCl. The van't Hoff factors for electrolytes will usually be less than the predicted values because of ion-pairing. That's where Na+ and Cl- attract and momentarily "look like" one particle before breaking back into two. Since the colligative properties are based on the number of particles in solution, it will appear that there are fewer particles. Therefore, a more realistic value of "i" for NaCl is 1.9 or 1.95. Still close to "2", but not quite.

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- billrussell42Lv 79 months ago
The boiling point is raised by 0.5 degrees Celsius for water with 29.2 grams of salt dissolved in each kg of water.

You have 2.94 g/114 g = 0.0258 g/g or 25.8 g/kg

0.5º per 29.2 g/kg x 25.8 g/kg = 0.44º

25.8 g/kg

- pisgahchemistLv 79 months agoReport
The Kb for water is 0.512 and that is in C/m where m is the molality, not the mass.

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- Roger the MoleLv 79 months ago
NaCl → Na{+} + Cl{-} [two ions]

(2.94 g NaCl) / (58.4430 g NaCl/mol) x (2 mol ions / 1 mol NaCl) / (0.114 kg) = 0.88255 mol/kg ions

(0.88255 m) x (0.512 °C/m) = 0.452°C increase

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- Anonymous9 months ago
1.21 gigawats.......

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