How many grams of Potassium Sulfate must be added to 6.00 x 10^-2 mL of water to form a 2.50m<-----molal solution
2007-07-03 12:54:48 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
chemistry...yuck..
google it and look for an equation for molality.
look for grams first.
2007-07-03 12:57:26 · answer #1 · answered by Casey A. 2 · 0⤊ 2⤋
Molality is defined as moles of solute / kg. of solvent
You can set up a ratio to solve this problem
calculate the grams in 2.5 moles of K2SO4. This would be 1 molal in 1 kg. of water
Not set the ratio to calculate the grams of K2SO4 in only 6.00 X 10E-2 grams of water.
2007-07-03 22:08:20 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋
2.50 molality solution suggests that there should be 2.5 moles of potassium sulfate to every 1 mole of water. Thus, you must determine the number of moles of water in 6.00x10-2 mL of water, then multiply by 2.5 to get the number of moles of potassium sulfate. Then you convert that number to grams by multiplying it by the molecular weight of potassium sulfate.
2007-07-03 20:24:52 · answer #3 · answered by Oh Snap! 2 · 0⤊ 1⤋
K2SO4 + HOH -----> 2KOH + HSO4(sulfuric acid)
?g 6 x 10^2ml 2.50m
molality = moles solute/kg solvent
You have the molality, and the volume of water.
1) convert volume of water to grams, then to moles using the density of water. 6 x 10^2ml x 1g/1mL x 1molH2O/18g H2O x 1000g/1kg. (I don't have a calculator handy, just solve this and you will get kg of water.
2.50m = moles of solute/kg of water from top problem.
after you get moles of K2SO4, convert the moles to grams by dividing by the total mass of K2SO4. I hope this helps :)
2007-07-03 22:19:33 · answer #4 · answered by Anonymous · 0⤊ 0⤋