269 mL of a 0.2089 M sodium carbonate (Na2CO3) solution is prepared and labelled solution A. Calculate the mass (in grams) of sodium carbonate (molar mass = 105.99 g mol-1) required to prepare Solution.
You should read your question and rewrite it appropriately.
To calculate the mass of sodium carbonate required to prepare the given solution, we need to use the equation:
Molarity (M) = moles of solute / volume of solution (in liters)
First, we need to convert the volume of the solution from milliliters (mL) to liters (L):
269 mL ÷ 1000 mL/L = 0.269 L
Now, rearrange the equation to solve for the moles of solute:
moles of solute = Molarity × volume of solution
moles of solute = 0.2089 M × 0.269 L
moles of solute = 0.0561671 mol
The molar mass of sodium carbonate (Na2CO3) is given as 105.99 g/mol. Multiply the moles of solute by the molar mass to find the mass of sodium carbonate required:
mass = moles of solute × molar mass
mass = 0.0561671 mol × 105.99 g/mol
mass = 5.95 grams
Therefore, the mass of sodium carbonate (Na2CO3) required to prepare Solution A is 5.95 grams.