How many atoms are in 56.7 grams Mg3P2?
N=mM⋅NAN=\frac{m}{M} \cdot N_AN=Mm⋅NA
where NNN - number of atoms; mmm - mass, g; MMM - molar mass, g/mol; NAN_ANA - Avogadro's number, NA≈6.022⋅1023 mol−1N_A \approx 6.022 \cdot 10^{23} \; mol^{-1}NA≈6.022⋅1023mol−1.
M(Mg3P2)=3⋅24.305+2⋅30.974=134.863 g/molM(Mg_3P_2)=3 \cdot 24.305+2 \cdot 30.974=134.863 \; g/molM(Mg3P2)=3⋅24.305+2⋅30.974=134.863g/mol
N=56.7134.863⋅6.022⋅1023≈2.53⋅1023N=\frac{56.7}{134.863} \cdot 6.022 \cdot 10^{23} \approx 2.53 \cdot 10^{23}N=134.86356.7⋅6.022⋅1023≈2.53⋅1023
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