Respuesta :
Answer:
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
Explanation:
Vapor pressure of water at 25 °C ,[tex]p^o= 23.8 mmHg[/tex]
Vapor pressure of the solution = [tex]p_s[/tex]
Number moles of water in 5.50% NaCl solution.In 100 gram of solution, 94.5 g of water is present.
[tex]n_1=\frac{94.5 g}{18 g/mol}=5.25 mol[/tex]
Number moles of NaCl in 5.50% NaCl solution.5.50 g of NaCl in 100 grams of solution.
[tex]n_2=\frac{5.50 g}{58.5 g/mol}=0.09401 mol [/tex]
Mole fraction of the solute = [tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\chi_2=\frac{0.09401 mol}{0.09401 mol+5.25 mol}=0.01759[/tex]
The relative lowering in vapor pressure of the solution with non volatile solute is equal to the mole fraction of solute in the solution:
[tex]\frac{p^o-p_s}{p^o}=\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\frac{23.8 mmHg - p_s}{23.8 mmHg}=0.01759[/tex]
[tex]p_s=23.38 mmHg[/tex]
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
23.38 mmHg is the vapour pressure.
The pressure enforced by the vapours when in the thermodynamical equilibrium state on the system is called the vapour pressure.
How to calculate the vapour pressure?
- Vapour pressure [tex](p^{\circ})[/tex] of water at [tex]25 ^{\circ} \rm C[/tex] = 23.8 mmHg
- Vapour pressure of aqueous solution = [tex](p_{s})[/tex]
Step 1: Moles of water in 5.50% NaCl solution when 100 grams of solution = 94.5 g of water
[tex]\begin{aligned}\rm n &= \dfrac{94.5 \;\rm g}{18\;\rm g/mol}\\\\\\&= 5.25\;\rm mol\end{aligned}[/tex]
Step 2: Moles of NaCl in 5.50% solution when 5.50 g NaCl in 100 grams solution then,
[tex]\begin{aligned}\rm n &= \dfrac{5.50\;\rm g}{58.5\;\rm g/mol}\\\\\\&= 0.09401\;\rm mol\end{aligned}[/tex]
Step 3: Calculate the mole fraction of the solute:
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}}\\\\\rm X_{2} &= \dfrac{0.09401\;\rm mol}{0.09401 + 5.25\;\rm mol}\\\\&= 0.0175\end{aligned}[/tex]
Step 4: Calculate the vapour pressure
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}} = \dfrac{p^{\circ}-p_{s}}{p^{\circ}}\\\\0.0175 &= \dfrac{23.38 \;\text{mmHg} -p_{s}}{23.38 \;\rm mmHg}\\\\&= 23.38\;\rm mmHg\end{aligned}[/tex]
Therefore, 23.38 mmHg is the vapour pressure.
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