To determine the amount of sodium dihydrogen phosphate required to prepare a 2M solution in 300 ml, we need to consider the molar concentration and the volume of the solution.
The formula for molarity (M) is:
Molarity (M) = (moles of solute) / (volume of solution in liters)
Rearranging the formula, we have:
moles of solute = Molarity (M) x volume of solution (in liters)
Given that we want to prepare a 2M solution in 300 ml (0.3 liters), we can calculate the moles of sodium dihydrogen phosphate required:
moles of solute = 2M x 0.3 L = 0.6 moles
Now, we need to convert the moles of sodium dihydrogen phosphate to grams using its molar mass. The molar mass of sodium dihydrogen phosphate (NaH2PO4) is:
Na: 22.99 g/mol H: 1.01 g/mol (x2 as there are two hydrogen atoms) P: 31.0 g/mol O: 16.00 g/mol (x4 as there are four oxygen atoms)
Summing up the molar masses:
22.99 g/mol + 1.01 g/mol (x2) + 31.0 g/mol + 16.00 g/mol (x4) = 119.98 g/mol
So, for 0.6 moles of sodium dihydrogen phosphate, the mass required is:
mass = moles x molar mass = 0.6 mol x 119.98 g/mol ≈ 71.99 grams
Therefore, approximately 71.99 grams of sodium dihydrogen phosphate would be required to prepare a 2M solution in 300 ml.