Of a non-volatile solute is dissolved in 365.0 g of water. the solute does not react with water nor dissociate in solution. assume that the resulting solution displays ideal raoult's law behaviour. at 70°c the vapour pressure of the solution is 231.16 torr. the vapour pressure of pure water at 70°c is 233.70 torr. calculate the molar mass of the solute (g/mol).

By using this formula of vapor pressure:
Pv(solu)= n Pv(water) 
when we have Pv(solu)=231.16 torr & Pv(water)= 233.7 torr
from this formula, we can get n (mole fraction of water) by substitution:
231.16 = n * 233.7
∴ n(mole fraction of water) = 0.99
so mole fraction of solution = 1 - 0.99 = 0.01
when no.of moles of water = mass weight / molar weight
                                             = 365g / 18g/mol = 20 moles
Total moles in solution =  moles of water / mole fraction of water
                                      = 20 / 0.99 =20.2
no. of moles of the solution= total moles in solution- moles of water
                                              = 20.2 - 20 = 0.2 moles
when we assumed the mass weight of the solution = 16 g (missing in your question should be given)
∴ molar mass = mass weight of solute / no. of moles of solute
                        = 16 g / 0.2 mol = 80 g/mol
 

The air is going to come out because of the low pressure.

Answer; carbon monoxide (co);