Common salt obtained from sea-water contains 95 % NaCl by mass. The approximate number of molecules present in 10.0 g of the salt is

$58.5 \mathrm{g}\text{ of }\mathrm{NaCl}\text{ contains }6.023\times {10}^{23}\text{ molecules. }$
10 g will contain 1.03 x 10²³ molecules of NaCI.
Since it contain 95 % NaCl, so total number of molecules
$=\frac{95}{100}\times 1.03\times {10}^{23}=0.98\times {10}^{23}\text{ or }{10}^{23}$