How many grams of CaCl2 (molar mass = 111.0 g/mol) are needed to prepare 4.44 L of 0.500 M CaCl2 solution?
To calculate the number of grams of CaCl2 needed to prepare the solution, we need to use the formula:
\(n = M \times V\)
Where:
\(n\) is the number of moles of CaCl2,
\(M\) is the molarity of the solution in mol/L, and
\(V\) is the volume of the solution in liters.
We can then use the molar mass of CaCl2 to convert moles to grams.
Let's calculate step-by-step:
Step 1: Calculate the number of moles of CaCl2
\(n = M \times V\)
\(n = 0.500 \, \text{mol/L} \times 4.44 \, \text{L}\)
\(n = 2.22 \, \text{mol}\)
Step 2: Convert moles to grams
To convert moles to grams, multiply the number of moles by the molar mass of CaCl2.
\(m = n \times \text{molar mass}\)
\(m = 2.22 \, \text{mol} \times 111.0 \, \text{g/mol}\)
\(m = 246.42 \, \text{g}\)
Therefore, 246.42 grams of CaCl2 are needed to prepare 4.44 L of 0.500 M CaCl2 solution.
To calculate the number of grams of CaCl2 needed, you can use the formula:
Mass = Molarity x Volume x Molar Mass
Given:
Molarity (M) = 0.500 M
Volume (V) = 4.44 L
Molar Mass (Mm) = 111.0 g/mol
Substituting these values into the formula gives:
Mass = 0.500 mol/L x 4.44 L x 111.0 g/mol
First, multiply 0.500 mol/L by 4.44 L:
= 2.22 mol
Then, multiply 2.22 mol by 111.0 g/mol:
Mass = 246.42 g
Therefore, you would need approximately 246.42 grams of CaCl2 to prepare 4.44 L of a 0.500 M CaCl2 solution.
How many moles do you need? M x L = moles.
Then moles CaCl2 = grams CaCl2/molar mass CaCl2.