what is the molarity of a solution prepared by dissolving 5.68 grams of NaOH in enough water to make 400. milliliters of solution?
How many grams of HCI is required to prepare 250 milliliters of a 0.158 M solution?
If 39 grams of benzene are dissolved in 100 grams of water, what is its boiling point? kb for water is 0.52 C/m
2007-08-01 03:06:52 · 2 answers · asked by Ashley T 1 in Science & Mathematics ➔ Chemistry
1) 5.68/40 x 1000/400
2) 250/1000 x 0.158 x 36.5
3) benzene doesn't dissolve in water at all. Bad question, I'm afraid!
2007-08-01 04:49:26 · answer #1 · answered by Gervald F 7 · 0⤊ 0⤋
5.68/40 = 0.142 moles NaOH
M = moles/L = 0.142/0.400 = 0.355M
v x 12 = 250 x 0.158
v = 3.29mL of HCl
density of HCl = 1.2g/mL
g/3.29 = 1.2
g = 3.95g of HCl
390/78 = 5.0 moles of benzene
5/.1 = 50m solution of benzene in water
delta T = 0.52 x 50
delta T= 26 degrees C
Therefore the bp is 126 C
Comment: benzene is not this soluble in water.
2007-08-01 11:49:32 · answer #2 · answered by Jabberwock 5 · 0⤊ 0⤋