What is the molality of a solution containing 12.1 g Na2SO4 dissolved in 2152.1 g of water?

Tolerance to high levels of salinity

Ithink that one is (c)

Use the universal gas law formula: pv = nrt now, the key word in this problem is stp, so at stp, temperature will always be 0 degrees celsius and pressure will always be 1.0 atm or 760 torr. given something is at stp, we already know the two variables pressure and temperature. the other thing we can find with the given information is moles. to find moles, just divide grams by molar mass of the element. 16.0g / 2(16.00)g/mol = 0.5 moles plug the values in to the equation pv = nrt (1.0 atm)( x ) = (0.5 )(0+273=273k) the r value is just a constant for when we use atm, liters, moles and kelvin. i'm assuming you want your answer in liters, but if you dont just move the decimal 3 to the right in your final answer. so when plugging these numbers in to your calculator and trying to find x, you get 11.20665 liters using significant figures the answer is 11.2 liters