Calculate weight of 6-022 - 1023 molecules of CaCO3-

1 mole of CaCO_3 contains 6.022× 10^23 molecules and 1 mole of CaCO_3 contains 1 mole of Ca , 1 mole of C a ...

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Standard XII

Physics

Measuring Mass

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Question

Calculate weight of $6.022 \times 10^{23}$ molecules of $C a C O_{3}$ .

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Solution

$1$ mole of $C a C O_{3}$ contains $6.022 \times 10^{23}$ molecules and $1$ mole of $C a C O_{3}$ contains $1$ mole of $C a$ , $1$ mole of $C$ and $3$ moles of $O x y g e n$ .
Therefore, weight of $6.022 \times 10^{23}$ molecules of $C a C O_{3}$
$=$ weight of $1$ mole of $C a C O_{3}$
$=$ (atomic mass of $C a$ ) $+$ (atomic mass of $C$ ) $+ 3$ (atomic mass of $O$ ) $+$
$= 40 + 12 + (3 \times 16)$
$= 100 g / m o l$

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Calculate weight of 6.022×1023 molecules of CaCO3.

1 mole of CaCO3 contains 6.022×1023 molecules and 1 mole of CaCO3 contains 1 mole of Ca, 1 mole of C and 3 moles of Oxygen.Therefore, weight of 6.022×1023 molecules of CaCO3 = weight of 1 mole of CaCO3 = (atomic mass of Ca) + (atomic mass of C) +3 (atomic mass of O) + =40+12+(3×16) =100g/mol

Calculate weight of $6.022 \times 10^{23}$ molecules of $C a C O_{3}$ .