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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

User Krsi
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1 Answer

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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,


(p^o-p_s)/(p^o)=(w_2M_1)/(w_1M_2)

where,


p^o = vapor pressure of pure solvent (water) = 760 torr


p_s = vapor pressure of solution = ?


w_2 = mass of solute (ethylene glycol) = 557.1 g


w_1 = mass of solvent (water) = 1000.0 g


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


M_2 = 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.


(760-p_s)/(760)=(557.1* 18.02)/(1000* 62.07)


p_s=637torr

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

User Redoubts
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