Question

common salt obtained from sea water contains 96% of NaCl by mass. the approximate no. of molecules present in 10.0 g of salt is

Answer 1

Let 100g of sea water contains 96g of sodium chloride
So 10g of water would have 9.6 g of sodium chloride
Molar mass of sodium chloride is 35.5+23=58.5g
No of moles in 9.6 g of sodium chloride is 9.6/58.5 moles
No. Of molecules present would be 9.6/58.5*Avogadro number

Answer 2

Answer: $0.96\times 10^{23}$ molecules of NaCl

Explanation:-

Given: mass of salt = 10 g

But as salt contains 96% of NaCl by mass, amount of NaCl=$\frac{96}{100}\times 10=9.6g$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}$

Mass of NaCl = 9.6 g

Molar mass of NaCl= 58.5 g/mol

Putting values in above equation, we get:

$\text{Moles}=\frac{9.6g}{58.5g/mol}=0.16mol$

According to Avogadro's law, 1 mole of every substance contains avogadro's number $(6.023\times 10^{23}$ of particles.

Thus 0.16 mol of NaCl contains=$\frac{6.023\times 10^{23}}{1}\times 0.16=0.96\times 10^{23}$ molecules of NaCl.