PrepRung
Sign inSign up free

JEE Main 2016 · Chemistry · Biomolecules and Polymers · hard

10.0 g of sucrose is dissolved in water to make 100 mL and placed in a 1 dm polarimeter tube. Specific rotations (in ^° mL g^-1 dm^-1) are sucrose…

Single correct option

10.0 g10.0\ \mathrm{g} of sucrose is dissolved in water to make 100 mL100\ \mathrm{mL} and placed in a 1 dm1\ \mathrm{dm} polarimeter tube. Specific rotations (in ∘ mL g−1 dm−1^\circ\ \mathrm{mL\,g^{-1}\,dm^{-1}}) are sucrose +66.5+66.5, glucose +52.5+52.5 and fructose −92.4-92.4. After complete hydrolysis (assume the volume stays 100 mL100\ \mathrm{mL}, and C12H22O11=342\mathrm{C_{12}H_{22}O_{11}}=342, C6H12O6=180\mathrm{C_6H_{12}O_6}=180), the observed rotation changes from +6.65∘+6.65^\circ to
  1. A. −1.05∘-1.05^\circ
  2. B. +1.05∘+1.05^\circ
  3. C. −1.99∘-1.99^\circ
  4. D. −2.10∘-2.10^\circ
Show answer and solution

Answer: (D) −2.10∘-2.10^\circ

Hydrolysis adds water: 342 g342\ \mathrm{g} sucrose gives 180+180=360 g180+180=360\ \mathrm{g} hexoses, so 10.010.0 g gives 10.0×360/342=10.5310.0\times360/342=10.53 g invert sugar, half glucose and half fructose (5.2635.263 g each, i.e. 0.05263 g mL−10.05263\ \mathrm{g\,mL^{-1}}). Observed rotation =∑[α] c l=(52.5−92.4)(0.05263)(1)=−2.10∘=\sum[\alpha]\,c\,l=(52.5-92.4)(0.05263)(1)=-2.10^\circ. The sign changes from ++ to −- (inversion). Ignoring the added water (taking 0.0500 g mL−10.0500\ \mathrm{g\,mL^{-1}} each) gives −1.99∘-1.99^\circ; halving the result gives −1.05∘-1.05^\circ.
Report a mistake

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake