(C) Glucose in its open-chain aldehyde form has $4$ chiral carbon atoms (at $C-2, C-3, C-4,$ and $C-5$).According to the formula for the number of ste

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(C) Glucose in its open-chain aldehyde form has $4$ chiral carbon atoms (at $C-2, C-3, C-4,$ and $C-5$).According to the formula for the number of stereoisomers,$N = 2^n$,where $n$ is the number of chiral centers.Here,$n = 4$,so the number of stereoisomers is $2^4 = 16$.