Using molarity and molality to described the solution. Make a solution
Calculate the percent concentration of 2.78g NaHCO3 dissolve in enough water to make 200.0ml of solution ?
To describe the solution, we need to calculate the molarity and molality first.
Molarity = moles of solute / liters of solution
First, we need to calculate the moles of NaHCO3:
molar mass of NaHCO3 = 23.0 + 1.0 + 12.0 + (3.0 x 16.0) = 84.0 g/mol
moles of NaHCO3 = (2.78 g) / (84.0 g/mol) = 0.0331 mol
Now, we need to calculate the liters of solution:
200.0 ml = 0.2 L
Molarity = 0.0331 mol / 0.2 L = 0.1655 M
Molality = moles of solute / mass of solvent (in kg)
Assuming the density of water is 1.00 g/mL, the mass of solvent in kg is:
mass of solvent = volume of solvent x density = 200.0 mL x 1.00 g/mL = 0.200 kg
Molality = 0.0331 mol / 0.200 kg = 0.1655 m
To calculate the percent concentration, we need to divide the mass of solute by the total mass of the solution and multiply by 100:
percent concentration = (mass of solute / total mass) x 100
mass of solute = 2.78 g
total mass = 200.0 mL x 1.00 g/mL + 2.78 g = 202.78 g
percent concentration = (2.78 g / 202.78 g) x 100 = 1.37%