Answer:
- [tex]n_{etOH}=0.11mol\ et OH[/tex]
- [tex]atoms_C=1.0x10^{23} atoms\ C[/tex]
Explanation:
Hello,
In this case, since ethanol has a molar mass of 46 g/mol, the moles in 5.0 g are:
[tex]n_{etOH}=5.0g\ etOH*\frac{1mol\ etOH}{46 g\ etOH} =0.11mol\ et OH[/tex]
Moreover, since the formula of glucose is C₆H₁₂O₆, its molar mass is 180 g/mol and six moles of carbon are in one mole of glucose (based on carbon's subscript), the atoms are computed by using the 6:1 mole ratio and the Avogadro's number as shown below:
[tex]atoms_C=5.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molC}{1molC_6H_{12}O_6} *\frac{6.022x10^{23}atoms\ C}{1molC} \\\\atoms_C=1.0x10^{23} atoms\ C[/tex]
Regards.
