How many grams of Na2CO3 can be found in a 273 mL sample of 0.5 M solution
To determine the number of grams of Na2CO3 in a 273 mL sample of a 0.5 M solution, we need to follow these steps:
Step 1: Convert the volume of the solution to liters.
To convert mL to liters, we divide the volume by 1000:
273 mL ÷ 1000 = 0.273 L
Step 2: Calculate the number of moles of Na2CO3 using the molarity (M) and volume in liters.
The formula to calculate the number of moles is:
moles = molarity × volume
moles = 0.5 M × 0.273 L
moles = 0.1365 mol
Step 3: Calculate the mass of Na2CO3 using the molar mass of Na2CO3.
The molar mass of Na2CO3 is:
Na: 22.99 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of Na2CO3 = 22.99 g/mol × 2 + 12.01 g/mol + 16.00 g/mol × 3
Molar mass of Na2CO3 = 46.00 g/mol + 12.01 g/mol + 48.00 g/mol
Molar mass of Na2CO3 = 106.01 g/mol
Step 4: Calculate the mass of Na2CO3 using the moles and molar mass.
The formula to calculate mass is:
mass = moles × molar mass
mass = 0.1365 mol × 106.01 g/mol
mass = 14.63 g
Therefore, there are 14.63 grams of Na2CO3 in a 273 mL sample of a 0.5 M solution.