How many atoms are in 167 g of calcium?

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How many atoms are in 167 g of calcium?

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Answer 1 · 2018-04-18T17:35:53

remind that mass=number of moles x molar mass and molar mass of Ca is 40
$167 = n \cdot 40$ $167=n*40$
$n = 4.175$ $n=4.175$
1 moles = $6.02 \times 10^{23}$ $6.02 × 10^23$ atoms
therefore 4.1752 moles of Ca= $4.175 \cdot 6.02 \times 10^{23}$ $4.175*6.02 × 10^23$ = $2.51335 \times 10^{24}$ $2.51335× 10^24$ atoms

Answer 2 · 2018-04-18T17:36:09

$m = 167 g$ $m=167g$
$M (C a) = 40.078 \frac{g}{m o l}$ $M(Ca)=40.078g/(mol)$

Explanation:

$n = \frac{m}{M} = 167 \frac{g}{40.078 \frac{g}{m o l}} = 4.167 m o l$ $n=m/M=167g/(40.078 g/(mol))=4.167 mol$
$n$ $n$ -numbers of mol

$N = n \cdot N_{a} = 4.167 m o l \cdot 6, 022 \times 10^{23}$ $N=n*N_a=4.167 mol*6,022 × 10^(23)$ (atoms)/(mol)
$= 25 \cdot 10^{23}$ $=25*10^(23)$ atoms

$N_{a}$ $N_a$ - Avogadro's constant $= 6, 022 \times 10^{23}$ $=6,022 × 10^(23)$ (atoms)/(mol)

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How many atoms are in 167 g of calcium?

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