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what is the mass of $1.56 \\times 10^{21}$ atoms of magnesium in grams?…

Question

what is the mass of $1.56 \times 10^{21}$ atoms of magnesium in grams?\
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$\bigcirc$ $4.72 \times 10^{-5}$\
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$\bigcirc$ $0.0630$\
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$\bigcirc$ $1.07 \times 10^{-4}$\
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$\bigcirc$ $0.142$\
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$\bigcirc$ none of the above

Explanation:

Step1: Recall Avogadro's number

Avogadro's number is \( 6.022\times 10^{23} \) atoms/mol. We need to find the moles of Mg first. The formula for moles from number of atoms is \( n=\frac{N}{N_A} \), where \( N \) is the number of atoms and \( N_A \) is Avogadro's number.
So, \( n=\frac{1.56\times 10^{21}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}} \)

Step2: Calculate moles of Mg

\( n=\frac{1.56\times 10^{21}}{6.022\times 10^{23}}\text{ mol}\approx 2.59\times 10^{-3}\text{ mol} \)

Step3: Recall molar mass of Mg

The molar mass of Mg is approximately \( 24.305 \) g/mol. The formula for mass is \( m = n\times M \), where \( M \) is molar mass.

Step4: Calculate mass of Mg

\( m=2.59\times 10^{-3}\text{ mol}\times24.305\text{ g/mol}\approx 0.0630\text{ g} \)

Answer:

0.0630 (corresponding to the option "0.0630")