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

mℓ

Explanation:

To solve this question one has to have in mind the values the quantum number can assume given a principal quantum number n.

ℓ angular  quantum numbe,can have the values 0, 1, 2,3 to n-1

and they correspond to the shape of the orbital in and have the symbol

0  ⇒   s orbital

1   ⇒   p orbial

2  ⇒   d orbital

for d orbitals ( l= 2 ) we can have five orientations in space  since  mℓ  can have the value which   range from - ℓ to + ℓ:

for l= 2  

mℓ = -2, -1, 0, 1, 2

which corresponds to the five orientations possible for d orbitals:

dz², dxy, dxz, dyz, dx²-y²

Hence it follows the two orbitals in our question differ in their quantum number  mℓ