The hydrocarbon (X) with molar mass 80 g mol ^-1 and 90% carbon has ____ degree of unsaturation.

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3 +4 marks

Solution & Explanation

### Related Formula For a hydrocarbon with the molecular formula mathrmC_x mathrmH_y, the Double Bond Equivalent (DBE) or Degree of Unsaturation (DU) is given by: textDU = x + 1 - fracy2 ### Core Logic 1. Calculate the mass of carbon in 1 mole of the hydrocarbon: textMass of Carbon = 80text g cdot 90\% = 72text g 2. Find the number of carbon atoms (x): x = frac7212 = 6 3. Find the mass and number of hydrogen atoms (y): textMass of Hydrogen = 80text g - 72text g = 8text g y = frac81 = 8 Thus, the molecular formula of hydrocarbon (X) is mathrmC_6mathrmH_8. 4. Calculate the degree of unsaturation: textDU = 6 + 1 - frac82 = 7 - 4 = 3 ### Pattern Recognition First, use the percentage composition and total molar mass to determine the exact number of carbon and hydrogen atoms. Once you have the molecular formula, plug it into the standard textDU = x + 1 - fracy2 equation to find the total number of rings and/or pi bonds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 16

Q72 jee_main_2024_31_jan_morning Purification Methods
'Adsorption' principle is used for which of the following purification method?
  • A. textExtraction
  • B. textChromatography
  • C. textDistillation
  • D. textSublimation

Solution

### Core Logic The fundamental principle used in chromatography (such as column chromatography or thin-layer chromatography) is adsorption. Different components of a mixture are adsorbed to different extents on an adsorbent stationary phase. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q80 jee_main_2024_31_jan_morning Purification Methods
Match List I with List II
LIST-I (Technique)LIST-II (Application)
A. DistillationI. Separation of glycerol from spent-lye
B. Fractional distillationII. Aniline - Water mixture
C. Steam distillationIII. Separation of crude oil fractions
D. Distillation under reduced pressureIV. Chloroform-Aniline
Choose the correct answer from the options given below:
  • A. textA-IV, B-I, C-II, D-III
  • B. textA-IV, B-III, C-II, D-I
  • C. textA-I, B-II, C-IV, D-III
  • D. textA-II, B-III, C-I, D-IV

Solution

### Core Logic (A) Simple distillation is used to separate a mixture of miscible liquids with a sufficiently large difference in their boiling points, such as chloroform and aniline. (A rightarrow IV) (B) Fractional distillation is used to separate crude oil fractions. (B rightarrow III) (C) Steam distillation is used to separate steam-volatile substances that are immiscible with water, like an aniline-water mixture. (C rightarrow II) (D) Distillation under reduced pressure is used for liquids that decompose at or below their normal boiling points, like separating glycerol from spent lye. (D rightarrow I) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

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