HClO₄ + KOH → KClO₄ + H₂O
HClO₄ - perchloric acid
KOH - potassium hydroxide
Answer:
0.209 mol/L
Explanation:
Given data
- Mass of copper(lI) sulfate (solute): 11.7 g
- Volume of solution: 350 mL = 0.350 L
The molar mass of copper(Il) sulfate is 159.61 g/mol. The moles corresponding to 11.7 grams are:
11.7 g × (1 mol/159.61 g) = 0.0733 mol
The molarity of copper(Il) sulfate is:
M = moles of solute / liters of solution
M = 0.0733 mol / 0.350 L
M = 0.209 mol/L
Explanation:
n=given mass ÷molar mass
make given mass become the subject of the formula by
multiplying the molar mass on both sides of the equation.
n=0.473mol
given mass=??
molar mass=48
therefore,given mass=n×molar mass
=0.473×48
=22.704grams
mass in grams is 22.704grams
