Question

What mass (in g) of urea (CO(NH2)2) in 100.0 g of water is needed to decrease the vapor pressure of water from 55.32 mmHg of pure water to 54.21 mmHg for the solution? The temperature is held constant at 40oC.

Answer 1

Answer: 6.7 g

Step-by-step 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,

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

where,

`p^o` = vapor pressure of pure solvent (water) = 55.32 mmHg

`p_s` = vapor pressure of solution = 54.21 mmHg

`w_2` = mass of solute (urea) = ? g

`w_1` = mass of solvent (water) = 100 g

`M_1` = molar mass of solvent (water) = 18 g/mole

`M_2` = molar mass of solute (urea) = 60 g/mole

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

`(55.32-54.21)/(55.32)=(x* 18)/(100* 60)`

`x=6.7g`

Therefore, 6.7 g of urea is needed in 100.0 g of water is needed to decrease the vapor pressure of water from 55.32 mmHg of pure water to 54.21 mmHg for the solution.