Calculate the mass of benzoic acid required to prepare 100cm3 each of 0.1m,0.25m, and 0.5m solutions
I assume your m should be Molarity, M.
mass=molartiy*molmass*volumeinLiters
mass=.1*molmassBenzoicAcid*.1m^3
so the mom mass is 122g/mole, so we have
mass=.1 Mole/liter*122g/mole*.1 Liter= 1.22 grams
check my thinking.
M =molar.volume. molar mass÷1000
314×220×1.28÷ 1000
= 0.535
To calculate the mass of benzoic acid required to prepare the given solutions, we need to use the formula:
Mass (g) = Molarity (mol/L) x Volume (L) x Molar mass (g/mol)
First, let's calculate the amount of benzoic acid required to prepare a 0.1 M solution.
Molarity (mol/L) = 0.1 mol/L
Volume (L) = 100 cm^3 ÷ 1000 = 0.1 L (since 1 cm^3 = 0.001 L)
Molar mass of benzoic acid (C7H6O2) = 122.12 g/mol
Mass (g) = 0.1 mol/L x 0.1 L x 122.12 g/mol
Mass = 1.2212 g
Therefore, to prepare a 0.1 M solution, you would need 1.2212 grams of benzoic acid.
Now let's move on to the 0.25 M solution.
Molarity (mol/L) = 0.25 mol/L
Volume (L) = 100 cm^3 ÷ 1000 = 0.1 L
Molar mass of benzoic acid (C7H6O2) = 122.12 g/mol
Mass (g) = 0.25 mol/L x 0.1 L x 122.12 g/mol
Mass = 3.052 g
Therefore, to prepare a 0.25 M solution, you would need 3.052 grams of benzoic acid.
Lastly, let's calculate the mass of benzoic acid required for a 0.5 M solution.
Molarity (mol/L) = 0.5 mol/L
Volume (L) = 100 cm^3 ÷ 1000 = 0.1 L
Molar mass of benzoic acid (C7H6O2) = 122.12 g/mol
Mass (g) = 0.5 mol/L x 0.1 L x 122.12 g/mol
Mass = 6.106 g
Therefore, to prepare a 0.5 M solution, you would need 6.106 grams of benzoic acid.