Question

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

Answer 1

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.