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Chemical equations and calculations

Question

6.023 x $10^{22}$ molecules are present in 10 g of a substance ‘Y’. The molarity of a solution containing 5 g of substance ‘Y’ in 2 L solution is ______ x $10^{- 3}$ .

Moderate

Solution

$6 .023 \times 10^{22}$ molecules in grams
$6 .023 \times 10^{23}$ molecules are present in 100 gms
Molecular mass of substance ‘Y’ = 100 gms
$M = \frac{5}{100 \times 2} = 0 .025$ $= 25 \times 10^{- 3}$

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