Question

How many moles of NaOH are present in 12.0mL of 0.110 M NaOH? Moles: _________

Answer 1

NaOH

M = molarity (mol/L)

`M\text{ = }\frac{mass\text{ of solute}}{\text{molecular mass of solute x volume of solution (L)}}=\frac{moles\text{ of solute}}{\text{volume of solution(L)}}`
`M=0.110\text{ }(mol)/(L)=\text{ }\frac{moles\text{ of NaOH}}{0.012L}`

12.0 ml = 0.012 L

`\text{moles of NaOH = 0.11 }(mol)/(L)x0.012L=1.32x10^{-3\text{ }}moles\text{ of NaOH}`

Answer: moles of NaOH = 1.32x10^-3 moles

`\text{moles NaOH = 1.32x10}^(-3)moles\text{ = 0.00132 moles NaOH}`