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The number of Glucose molecules present in 10 ml of decimolar solution is

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6.0 \times 10^{21}

6.0 \times 10^{22}

6.0 \times 10^{20}

6.0 \times 10^{19}

answer is A.

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Detailed Solution

M = \frac{n}{V_{(ml)}} \times 1000

$n = \frac{M \times V}{10^{3}} = \frac{0.1 \times 10}{10^{3}} = 10^{- 3}$

$No .of molecules = n \times 6.023 \times 10^{23}$

$= 10^{- 3} \times 6.023 \times 10^{23}$

$= 6.023 \times 10^{20}$

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