Question

How many moles of NaCl are required to prepare 2.90 L of 1.8 M NaCl?

Answer 1

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

Answer 2

Answer: 5.22 moles

Step-by-step explanation: Molarity of solution is defined as the number of moles of solute dissolved per liter of the solution.

`Molarity=\frac{\text{no of moles of NaCl}}{\text{Volume in L}}`

Given : Molarity of NaCl= 1.8 M

Volume of solution in liters = 2.90

Thus putting in the values , we get

`1.8M=\frac{\text{no of moles of NaCl}}{2.90L}`

`{\text {number of moles}=5.22 moles`