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In nature,abundance of $^1H_1$ and $^2H_1$ are 99.985 and 0.015 percent respectively and water obtained from it contains $H_2O$ and $D_2O$.The number of $D_2O$ molecules in 400mL water (if density is assumed to be 1.0g/mL) will be :

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

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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.
answered Oct 25, 2013 by sreemathi.v
edited Mar 22, 2014 by sharmaaparna1

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