Answer:
a) 0.5198 mol/L
b) 0.00811 mol KOH
Explanation:
a) M(KOH) = 39+16+1= 56 g/mol
14.555g* 1 mol/56 g = 14.555/56 mol KOH in 500.0 mL solution
14.555/56 mol KOH ---- 0.5000 L solution
x mol KOH ----- 1 L solution
x = (14.555/56)mol * 1L/0.5000 L = 0.5198 mol/L
b) If the student pours out a 15.6 mL sample of this solution, how many moles of sodium hydroxide? (if we talk about KOH)
does the student have in the sample
15.6 mL = 0.0156 L
0.5198 mol/L * 0.0156 L = 0.00811 mol KOH