What mass of Na2CO3 is required to prepare 607.0 mL of 0.250 M solution? Use a molar mass with at least as many significant figures as the data given.
____g?
How many moles do you need? That will be M x L = ? Na2CO3.
Then moles = grams/molar mass. Solve for grams.
To find the mass of Na2CO3 required to prepare the solution, we need to use the formula:
Mass = Volume x Molarity x molar mass
First, let's determine the molar mass of Na2CO3 (sodium carbonate):
Na (sodium) has a molar mass of 22.99 g/mol.
C (carbon) has a molar mass of 12.01 g/mol.
O (oxygen) has a molar mass of 16.00 g/mol.
To calculate the molar mass of Na2CO3:
2(Na) + 1(C) + 3(O) = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol)
= 45.98 g/mol + 12.01 g/mol + 48.00 g/mol
= 105.99 g/mol
Next, substitute the given values into the formula:
Mass = Volume x Molarity x molar mass
= 607.0 mL x 0.250 M x 105.99 g/mol
There is a unit conversion we need to perform before proceeding with the calculation:
1 L = 1000 mL
Converting the volume into liters:
607.0 mL x (1 L / 1000 mL) = 0.607 L
Now, we can find the mass:
Mass = 0.607 L x 0.250 M x 105.99 g/mol
= 0.15275 mol x 105.99 g/mol
= 16.185 g
Therefore, the mass of Na2CO3 required to prepare 607.0 mL of 0.250 M solution is approximately 16.185 grams (rounded to four significant figures).