12.8% KOH is 12 g KOH/100 g solution; therefore, we could prepare
100g x 12.0/12.8 = 93.75 g soln.
Of the 93.75 g soln, 12.0 g would be KOH; therefore 93.75-12.0 = 81.75 g water + 12.0 g KOH.
12.0 g KOH is how many moles?
12/molarmass = 12/56.1 = 0.214 moles.
M = moles/L. Substitute to obtain
0.29M = 0.214/L and
L = 0.214/0.29 = 0.738 L which you can convert to mL.
I fyou had only 12g of KOH remaining in a bottle, how many ml of 12.8%(w/v) solution could you prepare? (in mL)
How many ml of 0.29 M solution could you prepare using 12 g of KOH?
2 answers
12.75 g less than the theoretically predicted mass of 25.69 g, what is the percent yield of this reaction?
1 answer