Question

How many grams of urea [(nh2)2co] must be added to 329 g of water to give a solution with a vapor pressure 2.35 mmhg less than that of pure water at 30°c? (the vapor pressure of water at 30°c is 31.8 mmhg.)?

Answer 1

Answer : 87.60 g of Urea.

Explanation : We need to use the mole fraction formula here;

$x_{urea} = x_{water} . P^{o} _{water} }$

So, we have values and substituting them we get, 29.45 = $x_{water}$ (31.8) 

Hence, $x_{water}$ = 0.926 (this is the mole fraction of water)

Therefore, mole fraction of urea = 1 - 0.926 = 0.074,

moles of water present = 329 g / 18.0 mol = 18.27 moles,

Now, we have 0.074 moles of urea / 0.926 moles of water = x moles of urea / 18.27 moles of water,

Therefore, x = 1.46 moles of urea.

And now, Mass of urea = moles of urea X molar mass of urea;

= 1.46 X 60.0 = 87.60 g 

Hence the mass of urea to be dissolved in 329 g of water will be 87.60 g