A 25.0 mm × 40.0 mm piece of gold foil is 0.25 mm thick. The density of gold is 19.32 g/cm. How many gold atoms are in the sheet ? (Atomic weight : Au= 197.0)
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Answer:1.47×10^22 atoms
Explanation:
1 cm= 10 mm
Therefore,
Volume of gold sheet= 25/10 × 40/10× 0.25/10= 2.5×4×0.025
Volume=1/4 cubic cm
Density= Mass/Volume
19.32=M/1V/4=4M
19.32=4M
Given mass (M)=4.83 grams
Number of moles=given mass(M)/Molar mass
N=4.83/197= 0.024 moles
Number of moles= given number of particles(x)/Avagadro Constant
0.024=x/6.022×10^23
X= number of particles= 6.022 × 10^23 × 4.83 = 1.47×10^22 atoms
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