How many gold atoms in a $carat$ $"carat"$ ?

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How many gold atoms in a $carat$ $"carat"$ ?

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Answer 1 · 2016-08-09T06:44:45

$1 carat$ $"1 carat"$ $=$ $=$ $0.007055 oz$ $"0.007055 oz"$ $=$ $=$ $200 \cdot m g$ $200*mg$ , so...........

Explanation:

$No. of moles$ $"No. of moles"$ $=$ $=$ $\frac{mass}{molar mass}$ $"mass"/"molar mass"$ $=$ $=$

$\frac{22 carat \times 200 \cdot m g \cdot {carat}^{- 1}}{196.97 \cdot g \cdot m o l^{- 1}}$ $("22 carat"xx200*mg*"carat"^-1)/(196.97*g*mol^-1)$ $=$ $=$ $? ? m o l$ $??mol$

And $number of gold atoms$ $"number of gold atoms"$ $=$ $=$

$\frac{22 carat \times 200 \times 10^{- 3} g \cdot {carat}^{- 1}}{196.97 \cdot g \cdot m o l^{- 1}} \times 6.022 \times 10^{23} \cdot m o l^{- 1}$ $("22 carat"xx200xx10^-3g*"carat"^-1)/(196.97*g*mol^-1)xx6.022xx10^23*mol^-1$ , which gives a result as a NOOMBER!

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