Question

Compute the number of ions present in 11.7g od sodium chloride

Answer 1

NacL -23+58.5=58.5g (Molarmass-sum of atomic masses)  

Nacl-23  

cl-35.5  

a 58.5g nacl contains 23 grams of na  

23 in 58.5g  

there fore xgrams of na is present in 11.7g  

x in 11.7g  

x/23 =11.7/58.5  

x= 23* (11.7/58.5)=4.60g =~0.2mole

Answer 2

$\huge\red{Assalamualykum}$

_______________


$\huge\purple{Given :-}$

Mass of NaCl =11.7 g

Molar mass of NaCl = 23+35=58g/mol


$\huge\orange{Required:- }$


Number of ions=?

$\huge\blue{Solution :- }$

Moles of NaCl =mass/ molar mass

Moles of NaCl =11.7/58

Moles of NaCl = 0.2mol

Now ........

Moles of NaCl = Number of molecules /Avagadro number

Number of molecules = moles×Avagadro number

Number of molecules =0.2 × 6.023×10^23

Number of molecules of NaCl = 1.2×10^23

Now ......

We know that in one molecule of NaCl 2 ions that is Na+ and Cl_ are present so .......


1.2 ×2 ×10^23

$\huge\red{2.4×10^23 Ions }$


_________________



$\huge\pink{INSHALLAH it will help U}$