Question

if 0.0203 g of a gas dissolves in 1.39 l of water at 1.02 atmospheres at what pressure (in atm) would you be able to dissolve 0.146 g in the same amount of water?

Answer 1

First, we need to get n1 (no.of moles of water ): when

mass of water = 0.0203 g and the volume = 1.39 L

∴ n1 = mass / molar mass of water

        = 0.0203g / 18 g/mol
        = 0.00113 moles

then we need to get n2 (no of moles of water) after the mass has changed:

when the mass of water = 0.146 g

n2 = mass / molar mass

     = 0.146g / 18 g/ mol
     = 0.008 moles

so by using the ideal gas formula and when the volume is not changed:

So, P1/n1 = P2/n2 

when we have P1 = 1.02 atm 

and n1= 0.00113 moles

and n2 = 0.008 moles 

so we solve for P2 and get the pressure

∴P2 = P1*n2 / n1

        =1.02 atm *0.008 moles / 0.00113 moles

       = 7.22 atm

∴the new pressure will be 7.22 atm