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Chemistry 11 Online
OpenStudy (aaronandyson):

No. of atoms in 19.7g of Gold is?

OpenStudy (aaronandyson):

@Abhisar

OpenStudy (abhisar):

Ok, first you need to calculate the no. of moles of gold in 19.7g of gold. Do you know how to do that?

OpenStudy (abhisar):

\(\sf No.~of~moles=\huge \frac{Give~mass}{Atomic~mass}\)

OpenStudy (aaronandyson):

Shouldn't it be atomic mass/given mass?

OpenStudy (abhisar):

No. Given mass/Atomic Mass

OpenStudy (aaronandyson):

so,thats 0.1 mole?

OpenStudy (abhisar):

Yes, now multiply this by Avogadro's constant to get the number of atoms. Because one mole of anything has no of atoms equivalent to Avogadro's constant

OpenStudy (aaronandyson):

6.022 into 10^23

OpenStudy (abhisar):

Yes, that's avogadro's constant

OpenStudy (abhisar):

The answer would be \(\sf 0.1 \times N_A\)

OpenStudy (abhisar):

\(\sf N_A\)= Avogadro's Constant

OpenStudy (aaronandyson):

so is it 6.022*10^22?

OpenStudy (abhisar):

Yes, now do you understand how to do this?

OpenStudy (aaronandyson):

Yes.

OpenStudy (abhisar):

OK, cool

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