It is sharing (i could be wrong so be cautious of your answer) <span />
You need to use the ideal gas law (PV=nRT) and solve for n. ((3.50atm•10.0L)/(0.0821(L•atm/mol•K)•304K) = n = 1.40 moles. 1 mole of Cl2 = 70.9 gm/mole. The mass would be 99.43 gm
The % mass/mass concentration of solute in the seawater sample is calculated as below
% mass = mass of the solute /mass of the solvent(sea water) x100
mass of the solute =1.295 g
mass of the solvent(sea water_) = 25.895 g
there the % mass = 1.295/25.895 x100 = 5.001 %
