Question

Need help super fast! what is the molality of a solution containing 10.0g na2so4 dissolved in 1000.0g of water

Answer 1

-Find the molecular weight of Na2SO4.
-Calculate the amount of moles of na2SO4 you have by multiplying this amount by the molecular weight.
-Divide the amount of moles by the liters of solution (in this case 1000g is roughly 1000mL = 1 liter)
-then you have the molarity

Answer 2

How do I calculate the number of moles of Na2SO4 by molecular weight? What number do I multiply by?

Answer 3

to calculate the molecular weight you need a periodic table ( http://www.ptable.com)
Sodium (Na) = 22.98976 g/mol
Sulfur (S) = 32.065 g/mol
Oxygen (O) = 15.9994 g/mol

you have 2 sodium, 1 sulfur and 4 oxygen
22.98976 *2 = 45.97952
32.065 *1 = 32.065
15.9994 *4 = 63.9976
-------- +
142.04212 g/mol

142.04212 gram = 1 mol
10 gram = 10/142.04212 = 0.07040 mol

then you divide 0.07040 mol by the amount of liters

\[Molarity (M) = \frac{ Mol }{ Liter }\]

Answer 4

sorry link doesn't work its http://www.ptable.com

Answer 5

What about the 1000g of water?

Answer 6

there are no L just g

Answer 7

1000 grams of water almost equals 1 liter. This depends on the temperature.
At room temperature (20C) 1000 gram water = 997.567 Liter

Most simple is 1000g water (1KG) = 1000mL water (1L)

Answer 8

sorry At room temperature (20C) 1000 gram water = 997.567 miliLiter :P not liter

Answer 9

But how will I know for sure? It doesn't say anything about temperature in the question.....

Answer 10

If the temperature of density is not given you just take 1000g water =1 liter

Answer 11

Ok so so far I know the number of moles but thats all I really understand... How do I figure out the rest?

Answer 12

\[Molarity = \frac{ mol }{ liter }\] you know mol and liter

\[\frac{ 0.07040 }{ 1 } = 0.07040M \] (M=mol/liter)