Respuesta :
How many moles of NaCl are required to prepare 2.90 L of 1.8 M NaCl?
moles = molarity/ liters or mol= M/L
mol NaCl= 1.8 M/ 2.90 L=0.62 mol NaCl
moles = molarity/ liters or mol= M/L
mol NaCl= 1.8 M/ 2.90 L=0.62 mol NaCl
Answer: 5.22 moles
Explanation: Molarity of solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{\text{no of moles of NaCl}}{\text{Volume in L}}[/tex]
Given : Molarity of NaCl= 1.8 M
Volume of solution in liters = 2.90
Thus putting in the values , we get
[tex]1.8M=\frac{\text{no of moles of NaCl}}{2.90L}[/tex]
[tex]{\text {number of moles}=5.22 moles[/tex]