What mass of calcuim chloride is needed to prepare 3.650L of 1.75 M solution?
To find the mass of calcium chloride needed, we can use the equation:
\(M = \frac{n}{V}\)
where M is the molarity of the solution, n is the number of moles, and V is the volume of the solution.
Rearranging the equation to solve for n, we have:
\(n = M \times V\)
Given that the molarity M is 1.75 M and the volume V is 3.650 L, we can substitute these values into the equation:
\(n = 1.75 \times 3.650\)
Calculating this, we find that n is equal to 6.3875 moles.
To find the mass of calcium chloride needed, we need to know the molar mass of calcium chloride (CaCl2). The molar mass of CaCl2 is equal to the sum of the atomic masses of calcium (Ca) and chlorine (Cl) multiplied by the number of atoms in the compound.
The atomic mass of calcium is 40.08 g/mol, and the atomic mass of chlorine is 35.45 g/mol. Since there are two chlorine atoms in calcium chloride, the molar mass of CaCl2 is:
\(2 \times 35.45 + 40.08 = 110.98 \, \text{g/mol}\).
Now, we can calculate the mass of calcium chloride needed by multiplying the number of moles (n) by the molar mass of CaCl2:
\(m = n \times \text{molar mass}\)
Substituting the values, we have:
\(m = 6.3875 \, \text{moles} \times 110.98 \, \text{g/mol}\)
Calculating this, we find that the mass of calcium chloride needed is approximately 708.7 grams.