## Chemistry: The Central Science (13th Edition)

$4.44\times10^{-2}\;mol\;C_6H_{12}O_6$
Since we know that the solution is 6.45% glucose by mass, we can easily find the mass of glucose (molar mass = 180.16 g/mol) in solution. $124.0\;g\times6.45\%=7.998\;g\;C_6H_{12}O_6$ Next, convert to moles. $\frac{7.998\;g\;C_6H_{12}O_6}{}(\frac{1\;mol}{180.16\;g})=4.44\times10^{-2}\;mol\;C_6H_{12}O_6$