Question #106020
What is the mass of 2.234 x 10²³ (2.234E23) atoms of sulfur? (S = 32.07 g)
Round to the nearest tenth (1 decimal places)
1
Expert's answer
2020-03-23T10:14:38-0400

The amount in moles of the sulfur sample could be determined via the following:


n(S)=NatomsNA;n(S)=\frac{N_{atoms}}{N_{A}};


On the other hand, the amount in moles could be determined via the following:


n(S)=mSMS;n(S)=\frac{m_{S}}{M_{S}};


After combining of the both expressions:


mSMS=NatomsNA,\frac{m_{S}}{M_{S}}=\frac{N_{atoms}}{N_{A}},


or


mS=NMSNS=2.234102332.07g/mol6.0221023mol1=11.90g11.9g.m_{S}=\frac{N*M_{S}}{N_{S}}=\frac{2.234*10^{23}*32.07g/mol}{6.022*10^{23}mol^{-1}}=11.90g \sim 11.9g.

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