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JEE Main 2021 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 24 February 2021, Shift 2, Chemistry Q57

Question 57 of 90 in this shift, Chemistry question 27 of 30, Section B.

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25∘25^\circC. After 9 h, the fraction of sucrose remaining is ff. The value of log⁡10(1f)\log_{10}\left(\frac{1}{f}\right) is __________ ×10−2\times 10^{-2}. (Rounded off to the nearest integer) [Assume : ln 10 =2.303= 2.303, ln 2 =0.693= 0.693]

Official answer

81

NTA final key.

Same idea in other shifts

Asked 13× in all
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  6. 3 Aug 2021, Shift 2 · Q57The inactivation process of virus is first order with respect to virus concentration and 2% of the virus was inactivated in the first one…MediumNumerical value
  7. 31 Aug 2021, Shift 1 · Q56For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is __________. (Integer…EasyNumerical value
  8. 22 Jul 2021, Shift 2 · Q57N2O5(g)→2NO2(g)+12O2(g)\mathrm{N_2O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2} O_{2(g)}} In the above first order reaction the initial concentration of…EasyNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.