Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Some Basic Concepts of Chemistry question

2021 · 17 Mar · Shift 2 · Q15
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Some Basic Concepts of Chemistry
  5. /2021 · 17 Mar · Shift 2 · Q15

Some Basic Concepts of Chemistry question

2021 · 17 Mar · Shift 2 · Q15

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
JEE Main 2021 (Online) 17th March Evening Shift Chemistry - Some Basic Concepts of Chemistry Question 148 English Consider the above reaction. The percentage yield of amide product is ‾\underline{\hspace{2cm}}​. (Round off to the Nearest Integer). (Given : Atomic mass : C : 12.0 u, H : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
Numerical answer
View written solutionFree

Correct answer: 77

The image/reaction details referred to by “Consider the above reaction” are not present in the prompt, so the reactants, amounts, and actual product data needed to compute percentage yield are missing.

To calculate percentage yield of an amide, we would need:

  1. Balanced reaction for amide formation.
  2. Given amounts of reactants to identify the limiting reagent.
  3. Actual amount of amide obtained.
  4. Then use % yield=actual yieldtheoretical yield×100\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100% yield=theoretical yieldactual yield​×100

Since these required numerical details are absent, the value cannot be independently derived from the provided text alone.

However, the stored correct answer is given as 77. In absence of the missing reaction data, I cannot verify it rigorously, but if the expected final integer is required from the provided record, then the answer is:

77\boxed{77}77​

PreviousNext

More from Some Basic Concepts of Chemistry

  • ​ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) (C3​H4​O) if the percentage yield is 64. (Round off to the Nearest Integer). [Given : Atomic masses : C : 12.0 u,…2021 · Numerical
  • A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n × 10 − 1, when n = ​. (Round off…2021 · Numerical
  • Complete combustion of 3g of ethane gives x × 1022 molecules of water. The value of x is ​. (Round off to the Nearest Integer). [Use : NA = 6.023 × 1023; Atomic masses in u : C : 12.0; O : 16.0; H :…2021 · Numerical
  • 10.0 mL of Na2​CO3​ solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings : 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL Based on these readings, and convention of titrimetric estimation…2021 · Numerical
  • 250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ​× 1021. (Nearest integer) (NA = 6.022 × 1023)2021 · Numerical
  • An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass = 180.0 g mol − 1) needed to meet this energy requirement is ​ g. (Use : Δ CH(glucose) = − 2700 kJ mol − 1)2021 · Numerical
  • 4g equimolar mixture of NaOH and Na2​CO3​ contains x g of NaOH and y g of Na2​CO3​. The value of x is ​ g. (Nearest integer)2021 · Numerical
  • If the concentration of glucose (C6​H12​O6​) in blood is 0.72 g L − 1, the molarity of glucose in blood is ​× 10 − 3 M. (Nearest integer) [Given : Atomic mass of C = 12, H = 1, O = 16 u]2021 · Numerical