Question

Common salt obtained from sea-water contains 95% NaCI by mass. The approximate number of molecules present in 10 g salt is:

• A. ${10}^{21}$
• B. ${10}^{22}$
• C. ${10}^{23}$
• D. ${10}^{24}$

Answer 1

Answer: (C)

${10}^{23}$

100 g salt contains 95g NaCl

Therefore, 10g salt contains 9.5g NaCl

no. of molecules =(moles of NaCl) $\times 6.02\times {10}^{23}$

=$\frac{9.5}{58.5}$ $\times 6.02\times {10}^{23}$ = 10${}^{23}$