Question

How many mL of a 0.375 M solution can be made from 35 g of calcium phosphate?

Answer 1

To make a 0.375 M solution from 35 g of calcium phosphate, you would need approximately 300.2 mL of the solution.

Step-by-step explanation:

The molarity (M) of a solution can be calculated using the formula:

M = moles/volume in liters

To find the number of moles of calcium phosphate, we first need to calculate its molecular weight. Calcium phosphate has the chemical formula Ca3(PO4)2, which means it contains 3 calcium (Ca) atoms, 2 phosphate (PO4) ions, and 6 oxygen (O) atoms. The atomic weight of calcium is 40.08 g/mol, phosphorus is 30.97 g/mol, and oxygen is 16.00 g/mol. Therefore, the molecular weight of calcium phosphate is

(3 * 40.08) + (2 * (30.97 + (4 * 16.00))) = 310.18 g/mol

Now, we can use the formula to calculate the moles of calcium phosphate:

moles = mass/molecular weight = 35 g/310.18 g/mol = 0.1127 mol

Finally, we can calculate the volume of the 0.375 M solution:

volume = moles/M = 0.1127 mol/0.375 M = 0.3002 L = 300.2 mL

Answer 2

300 mL

Step-by-step explanation:

the unit formula of calcium phosphate is Ca3(PO4)2

molar mass of Ca3(PO4)2 = (3×40 + 2×31 + 8×16) g/mol = 310 g/mol

n = m/M = 35 g/(310 g/mol)

c = n/V

V = n/c = [35 g/(310 g/mol)]/0.375 mol/L

V = 0.30 L = 300 mL