the balanced equation for the acid base reaction is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
the number of NaOH moles is - 0.139 mol/L x 0.0154 L = 0.00214 mol
the number of NaOH moles reacted = number of HCl moles reacted
therefore number of HCl moles reacted = 0.00214 mol
volume of HCl containing 0.00214 mol - 25.0 mL
number of HCl moles in 25.0 mL - 0.00214 mol
therefore number of HCl moles in 1000 mL - 0.00214 mol / 25.0 mL x 1000 mL/L
molarity of HCl is 0.0856 mol/L
concentration of HCl is 0.0856 M
((mass solute)/(mass solution)) * 100% =35%
mass CaCl2=0.35*352.5=123.4 g CaCl2
When global winds, ocean currents, and other forces move freely across the earths surface.
Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
