What weight of Ca(OH)2 is necessary to prepare 3,500ml of 0.7M of Ca(OH)2 solution?
you want to have 3.5*.7 moles Ca(OH)2
mass needed= .7(3.5)*formulamass Ca(OH2)
To determine the weight of Ca(OH)2 necessary to prepare the given solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (L)
First, let's determine the number of moles of Ca(OH)2 needed. We rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) × volume of solution (L)
Molarity (M) = 0.7 M (given)
Volume of solution = 3,500 ml = 3.5 L (since 1 L = 1,000 ml)
Now substitute these values into the formula:
moles of solute = 0.7 M × 3.5 L = 2.45 moles
To calculate the weight of Ca(OH)2, we'll use the molar mass of Ca(OH)2, which is 74.09 g/mol (40.08 g/mol for calcium + 2 * 1.01 g/mol for hydrogen + 16.00 g/mol for oxygen).
weight of Ca(OH)2 = moles of solute × molar mass of Ca(OH)2
= 2.45 moles × 74.09 g/mol
= 181.16 g (rounded to two decimal places)
Therefore, to prepare 3,500 ml of 0.7M Ca(OH)2 solution, you will need 181.16 grams of Ca(OH)2.