Answer
0.123 M $ C _{6}H_{12}0_{6} $
Work Step by Step
$\frac{ 28.33 grams C _{6}H_{12}0_{6} }{1.28 L}\times\frac{1 molC _{6}H_{12}0_{6} }{180.16 gramsC _{6}H_{12}0_{6} } $
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