Calculate the vapor pressure lowering of a solution at 100.0 °C that contains 557.1 g of ethylene glycol (molar mass g/mol). The vapor pressure of pure water at T00.0 °C is 760 torr. (2 points) 62.07 g/mol) in 1000.0 g of water (H20 molar mass 18.02 A) 106 torr B) 0.756 torr C) 760 torr (D)186 torr E) none of these

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kobenhavn kobenhavn · Answer 1 · 2019-09-09T07:10:23+02:00

Answer : E) none of these.

The vapor pressure of solution is 637 torr .

Solution :

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}=\frac{w_2M_1}{w_1M_2}[/tex]

where,

[tex]p^o[/tex] = vapor pressure of pure solvent (water) = 760 torr

[tex]p_s[/tex] = vapor pressure of solution = ?

[tex]w_2[/tex] = mass of solute (ethylene glycol) = 557.1 g

[tex]w_1[/tex] = mass of solvent (water) = 1000.0 g

[tex]M_1[/tex] = molar mass of solvent (water) = 18.02 g/mole

[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62.07 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

[tex]\frac{760-p_s}{760}=\frac{557.1\times 18.02}{1000\times 62.07}[/tex]

[tex]p_s=637torr[/tex]

Therefore, the vapor pressure of solution is, 637 torr.