Question
The equilibrium constant of mutarotation of $\alpha$-D-glucose to $\beta$-D-glucose is 1.8 . What per cent of the $\alpha$-form remains under equilibrium?(a) 35.7(b) 64.3(c) 55.6(d) 44.4
Step 1
Since the equilibrium constant \( K \) for the mutarotation is given as 1.8, we can express the equilibrium concentrations in terms of \( x \). Show more…
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The equilibrium constant for the mutarotation, $\alpha$ -D-glucose $\rightleftharpoons \beta-D$ glucose is $1.8 .$ What per cent of the $\alpha$ -form remains under equilibrium? (a) $35.7$ (b) $64.3$ (c) $55.6$ (d) $44.4$
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The equilibrium constant of mutarotation of $\alpha-\mathrm{D}$ glucose to $\beta$ -D-glucose is $1.8 .$ What per cent of the $\alpha$ -form remains under equilibrium: (a) $35.7$ (b) $64.3$ (c) $55.6$ (d) $44.4$
In an aqueous solution of D-glucose the percentages of $\alpha$ - and $\beta$ -anomer at the equilibrium condition are respectively (a) $20 \%$ and $80 \%$ (b) $80 \%$ and $20 \%$ (c) $36 \%$ and $64 \%$ (d) $64 \%$ and $36 \%$
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