calculate the mass of impure KOH needed to make up 1.20 L of 0.60 mol/L KOH(aq)? Assume the impure KOH is 84% KOH by mass and 16% water.
try this;
n = Mv = 0.6x1.2 = 0.72moles
so, 0.72moles is the total mole i.e. KOH and water. Since the impure KOH is 84%, then the mole for KOH is;
0.84 x 0.72 = 0.605moles
so, m = nMr = 0.605x(molar mass of KOH).
To calculate the mass of impure KOH needed, we need to consider the percentage composition of KOH in the impure sample. In this case, the impure KOH is 84% KOH by mass and 16% water.
First, we need to determine the mass of KOH in the impure sample. Assuming we have 100 grams of the impure KOH, 84 grams would be KOH and 16 grams would be water.
Next, we need to convert the given concentration of KOH(aq) into moles.
Concentration (mol/L) = Moles/Volume (L)
Here, the given concentration is 0.60 mol/L and the volume is 1.20 L. Rearranging the equation, we have:
Moles = Concentration x Volume
Moles = 0.60 mol/L x 1.20 L
Moles = 0.72 mol
Since there is a 1:1 stoichiometric ratio between KOH and moles of KOH, we know that the moles of impure KOH should also be 0.72 mol.
Now, we can calculate the mass of impure KOH needed using the percentage composition information:
Mass of KOH = (Moles of KOH / 100) x Mass of the impure KOH
Since the impure KOH is 84% KOH by mass, we have:
Mass of KOH = (0.72 mol / 100) x 84 grams
Mass of KOH = 0.72 x 0.84
Mass of KOH = 0.6048 grams
Therefore, you would need approximately 0.605 grams of impure KOH to make up 1.20 L of 0.60 mol/L KOH(aq).