Question

how many
Na+ ions are present in hundred ml of 0.25 molarity of NaCl solution​

Answer 1

the formula for this is : (x/GMW)/volume*100

So, it is : (x/58.5)/100 =0.25

=> x/58.5 =25

i.e : that number of NaCl present in 100 ml of solution are 25.

So after complete dissasociation Na+ ions formed are 25.

“`````````````` answer is 25 Na+ ions are present.````

hope it helps you ❤️

Answer 2

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