12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68 mole(s) of CsF(s) in 1.00 kg of water? Assume that Raoult's law applies.

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of = 1.00 kg = 1000 g

Moles of = 3.68 mole

Molar mass of = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of .

Now we have to calculate the mole fraction of

Now we have to partial pressure of solution.

According to the Raoult's law,

where,

= vapor pressure of solution

= vapor pressure of water = 0.692 atm

= mole fraction of water = 0.938

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.

weight of any object is given by

w = m*g

where m is mass of the object and g is the acceleration due to gravity.

for earth the value of g is 9.8 m/s^2 so

w=117.6 n m=12 g=?

117.6=12g => g=9.8 m/s^2

which is the force of gravity on earth.