What is the molarity of a solution made with 64 grams of sodium hydroxide in 4 liters of water

The molarity of the solution: 0,27M

Explanation:

First , we calculate the weight of 1 mol of NaCl:

Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol

58,5 g---1 mol NaCl

64 gx= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl

A solution molar--> moles of solute in 1 L of solution:

4 L1,09 mol NaCl

1Lx0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M