Respuesta :
The boiling point of water generally increases as the amount of impurities (which a solute like KCl technically can be thought of) dissolved increases. This relation can be quantified using the equation,
[tex]\Delta T_b = i \times K_b \times m[/tex]
where [tex]\Delta{T}_{b}[/tex] is the change in the water's boiling point (normally taken to be 100 °C), [tex]i[/tex] is the Van 't Hoff factor (the number of particles a single formula unit of the solute dissociates into in water), [tex]K_b[/tex] is the boiling point elevation constant, and [tex]m[/tex] is the molality (moles of solute/kilogram(s) of solvent) of the solution.
We are forming a solution by dissolving KCl in water. KCl is an electrolyte that, in water, will dissociate into K⁺ and Cl⁻ ions. So, for every formula unit, KCl, we obtain two particles. Thus, the Van 't Hoff factor, or [tex]i[/tex], will be 2.
The molality of the solution can be calculated by dividing the number of moles of KCl by the mass of water in kilograms. Since we have 1.00 kg of water, we would be dividing 0.75 mol KCl by 1, giving us a molality (m) of 0.75 m.
We aren't provided the boiling point elevation constant for water. Several authoritative sources give the value 0.512 °C/m, so we will adopt that as our [tex]K_b[/tex].
Note: m = mol/kg as used in this problem.
Plugging everything in,
[tex]\Delta T_b = i \times K_b \times m \\\Delta T_b = 2 \times 0.512 \text{ } \frac{^oC}{mol/kg} \times 0.75 \text{ } \frac{mol}{kg} \\\Delta T_b = 0.768 \text{ } \mathrm{ ^oC}[/tex]
As you can see, our change in boiling point is positive (the boiling point is elevated), and it is also quite modest. Taking 100 °C to be the boiling point of pure water, the boiling point of our solution would be 100 ⁰C + 0.768 ⁰C, or 100.768 ⁰C.
If we are considering significant figures, then we must give our answer to two significant figures (since 0.75 has two sig figs). We can regard the boiling point of water (100 ⁰C) as a defined value. Since our final answer is a sum, the boiling point of our solution to two significant figures would be 100.77 ⁰C.
The boiling point of a solution formed will be "100.765°C".
Given:
- Mol = 0.75
- Mass = 1.00 kg
We know,
- Boiling point constant, Kb = 0.51
The molality of the solution will be:
= [tex]\frac{Mole}{Mass}[/tex]
= [tex]\frac{0.75}{1}[/tex]
= [tex]0.75 \ m[/tex]
Now,
→ [tex]T_{solution}-T_{water} = Kb\times m\times i[/tex]
By putting the values, we get
[tex]= 0.51\times 0.75\times 2[/tex]
[tex]= 0.765[/tex]
- Boiling point of water = 100°C
hence,
Solution's boiling point will be:
→ [tex]T_{solution} = 100+0.765[/tex]
[tex]= 100.765^{\circ} C[/tex]
Thus the above approach is right.
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