Question

Common salt obtained from sea-water contains $96\%NaCl$ by mass. The approximate number of molecules of $NaCl$ present in $10.0g$ of the common salt is:
$(At.wt.Na=23g)$

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

Answer 1

The correct option is C ${10}^{23}$

$\text{Mass of}NaCl=10\times 0.96=9.6gm$
$\text{moles of}NaCl=\frac{9.6}{58.5}mol$

$\text{Number of molecules}=\text{number of moles}\times N_A=\frac{9.6}{58.5}\times 6.023\times {10}^{23}\simeq {10}^{23}$