how many grans of Ca(OH) are needed to prepare 800.0 ml solution of 0.15M concentration? molar mass of Ca(OH) is 74g/mole
Hi faye,
Alright so we want to prepare a 800ml solution of 0.15M concentration.
So first we calculate the amount of Ca(OH)2 **NOTE: SHOULD NOT BE CA(OH) COS +2 CHARGE OF CA2+ IS NOT BALANCED! CA(OH)2 IS A SOLID AND NOT AN ION, HENCE THE COMPOUND CANNOT HAVE AN OVERALL CHARGE. **
Amount (mol) of Ca(OH)2 = M (mol/dm^3) X L (dm^3) = 0.15 X 800.0/1000 (must convert to appropriate units)
Then mass (g) = molar mass (g/mol) X amount (mol) giving you your mass of Ca(OH)2 needed.
Hope I helped! (:
-J
Oops, sorry, my apologies for typing in caps. Was just trying to point that out. :)
...and the answer should only be to 2 sig figs.
To determine the number of grams of Ca(OH)2 needed to prepare the solution, we'll use the formula:
Grams of solute = (Molarity) x (Volume in liters) x (Molar mass)
First, let's convert the volume from milliliters to liters:
Volume = 800.0 ml / 1000 = 0.8 L
Now, we can calculate the grams of Ca(OH)2:
Grams of Ca(OH)2 = (0.15 M) x (0.8 L) x (74 g/mol)
Multiplying these values, we get:
Grams of Ca(OH)2 = 8.88 g
Therefore, you would need 8.88 grams of Ca(OH)2 to prepare an 800.0 ml solution with a molar concentration of 0.15 M.