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?

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