100 ml aqueous solution (density $\frac{5}{3}$ gm/ml) contains 3 40 % by weight $\mathrm{NaOH}$. The number of molecules of $\mathrm{NaOH}$ dissolved in the above solution is: (Use $N_A=6\times {10}^{23}$)

mass of $\mathrm{NaOH}=\frac{40}{100}\times \frac{5}{3}\times 100=\frac{200}{3}$

now, moles of $\mathrm{NaOH}=$ Given weight of NaOH/Molar mass of NaOH

$=\frac{200}{3\times 40}=\frac{5}{3}$

Now number of molecules of NaOH= number of moles $\times N_A$

$=\frac{5}{3}\times 6\times {10}^{23}={10}^{24}$