Respuesta :
Answer:
0.808 M
Explanation:
Using Raoult's Law
[tex]\frac{P_s}{Pi}= x_i[/tex]
where:
[tex]P_s[/tex] = vapor pressure of sea water( solution) = 23.09 mmHg
[tex]P_i[/tex] = vapor pressure of pure water (solute) = 23.76 mmHg
[tex]x_i[/tex] = mole fraction of water
∴
[tex]\frac{23.09}{23.76}= x_i[/tex]
[tex]x_i = 0.9718[/tex]
[tex]x_i+x_2=1[/tex]
[tex]x_2 = 1- x_i[/tex]
[tex]x_2 = 1- 0.9718[/tex]
[tex]x_2 = 0.0282[/tex]
[tex]x_i = \frac{n_i}{n_i+n_2}[/tex] ------ equation (1)
[tex]x_2 = \frac{n_2}{n_i+n_2}[/tex] ------ equation (2)
where; [tex](n_2) =[/tex] number of moles of sea water
[tex](n_i) =[/tex] number of moles of pure water
equating above equation 1 and 2; we have :
[tex]\frac{n_2}{n_i}[/tex][tex]= \frac{0.0282}{0.9178}[/tex]
[tex]= 0.02901[/tex]
NOW, Molarity = [tex]\frac{moles of sea water}{mass of pure water }*1000[/tex]
[tex]= \frac{0.02901}{18}*1000[/tex]
[tex]= 0.001616*1000[/tex]
[tex]= 1.616 M[/tex]
As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have [tex]\frac{1.616}{2} =0.808 M[/tex]
Answer:
Molal concentration of NaCl in Seawater = 1.6 moles of NaCl/kg of pure water!!!
Explanation:
Let the mole fraction of pure water be X, that of NaCl = 1 - X
We obtain the mole fraction of solvent from Raoult's law relation
Pressure of solution, P⁰ₛₐₗₜ = Vapour pressure of solvent, P⁰ₕ₂₀ × mole fraction of solvent, X
23.09 = 23.76 × X
X = 0.972
Using a mole basis of 1 mole,
Number of moles of pure water = 0.972
number of moles of NaCl in the seawater sampler = 1 - 0.972 = 0.028 moles
Molal concentration = number of moles of solute/mass of solvent in kg
Mass of solvent = number of moles × molar mass of pure water = 0.972 × 18 = 17.496g = 0.0175 kg
Molal concentration = 0.028/0.0175 = 1.6 moles of NaCl/kg of pure water!!!
Hope this Helps!!!!
