Question

Common salt obtained from sea-water contains $96\%$ $NaCl$ by mass. The approximate number of molecules present in $10.0g$ of the salt is :

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

Answer 1

Answer: (C)

${10}^{23}$

Mass of NaCl $=10\times 0.96=$ $9.6$ g
Moles of NaCl = $\frac{9.6}{58.5}$
Number of molecules = $\frac{9.6}{58.5}\times 6.023\times {10}^{23}\approx {10}^{23}$.