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Sagot :
Answer: The vapor pressure of the solution at [tex]29^0C[/tex] is 29.86 mm Hg
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor = 1 (for non electrolytes)
[tex]x_2[/tex] = mole fraction of solute
=[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]
Given : 9 moles of [tex]NH_3[/tex] are dissolved in 50 L or 50000 ml of water
mass of water = [tex]density\times volume = 1g/ml\times 50000ml=50000g[/tex]
moles of solvent (water) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50000g}{18g/mol}=2778moles[/tex]
Total moles = moles of solute + moles of solvent = 9 mol + 2778 mol = 2787
[tex]x_2[/tex] = mole fraction of solute
=[tex]\frac{9}{2787}=3.2\times 10^{-3}[/tex]
[tex]\frac{29.96-p_s}{29.96}=1\times 3.2\times 10^{-3}[/tex]
[tex]p_s=29.86mmHg[/tex]
Thus the vapor pressure of the solution at [tex]29^0C[/tex] is 29.86 mm Hg
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