$\begin{array}{1 1}(a)\;10^{21}&(b)\;2\times 10^{21}\\(c)\;3\times 10^{21}&(d)\;4\times 10^{21}\end{array}$

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Total number of molecules =$\large\frac{v\times cl}{18}$$\times 6.023\times 10^{23}$

(assuming $\large\frac{18\times 99.985+20\times 0.015}{100}$$\approx 18)$

No of $D_2O$ molecules $=\large\frac{400\times 1}{18}$$\times 6.023\times 10^{23}$$\times \large\frac{0.015}{100}$

$\Rightarrow 2.01\times 10^{21}$

Hence (b) is the correct option.

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