Skip to main content
DAY 3
23.4.20

GOOD MORNING BOYS,
Today we will have assesment of topics covered in the form of MCQ through google form, after that we will cover the topics of limiting reagent and methods of expressing concentration.

FIRST ATTEMPT THE MCQ .




  LEARNING OUTCOMES;
Students will be able to
  • Define limiting reagent
  • Apply concept
  • learn different methods of expressing concentration.

Limiting Reagent:
It is the reactant which  gets consumed first or limits the amount of product formed.
For a balanced reaction reaction:
                                    A +B →  C + D
B would be a limiting reagent if nA /n> nB/nA
Similarly, A is a limiting reagent if nA /n< nB/nA
Concentration of the solutions
Mass Percent:
  • It is the mass of the solute in grams per 100 grams of the solution.
  • mass percent formula
Volume Percent:
  • It is the volume of the solute  per 100 units of the volume of solution.
  • volume percent of solute
Parts per million ( ppm):
  • It is the amount of the solute in gram per million (106) gram of the solution.
  • parts per million formula
Mole fraction:
  • It is the ratio of the moles of one component of the solution to the total number of moles of solution
  • Total mole fraction of all the components of a solution is equal to 1.
Molarity (M):
  •  It is the number of moles of solute dissolved per litre (dm3) of the solution.
  • molarity calculation formula
Molality (m):
  • It is the number of moles of solute present in 1 kg of solvent.
  • molality calculation formula
Note: Why m is not affected by change in temperature?
Since it is dependent on mass of solute and mass do not change to small shift in temperature due to close packing of solid atoms therefore it is independent of temperature change.
video to revise mole concept, percentage composition and empirical formula




Numerical practice :
What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 liters?
Solution:
            = 0.94 M NaCl
What is the molarity of a solution containing 0.46 mole of solute in 2.0 kg water?

Solution:
= 0.23 m
THATS ALL FOR THE DAY, WE WILL CONTINUE WITH NUMERICAL PRACTICE IN OUR NEXT THEORY CLASS.
STAY SAFE AND TAKE CARE.
MARK YOUR ATTENDANCE ON GOOGLE FORM AS WELL AS COMMENT SECTION OF BLOGGER.

Comments

  1. Replies
    1. ma'am
      the first question has two options correct if we consider it to be an oxygen molecule(O2) however the answer is correct when we take oxygen molecule to be O. so this was a multi correct question but only one option could be chosen.

      Delete
    2. Exactly ma'am...the answer should be 14g N2 ...where I differ is that,only this option should be correct..as taking the mass of oxygen molecule as that of O instead of O2 is meaningless...

      Delete
    3. Oxygen molecule will always be O2

      Delete
  2. Anshuman jaison of class 11D is present

    ReplyDelete
  3. Good Morning Ma'am
    Shaun Lawrence of 11-D
    Present

    ReplyDelete
  4. good morning ma'am
    sami ansari 11-D
    present

    ReplyDelete
  5. Jaskeerat Singh (11-D) - Present

    ReplyDelete
  6. Good morning ma'am,
    Celestine Saji

    ReplyDelete
  7. Good morning ma'am
    Avikam Gupta
    11-D

    ReplyDelete
  8. Good Morning,
    This is Joseph James Nedumpara of XI D.

    ReplyDelete
  9. GOOD MORNING MA'AM
    TEGHVEER SINGH
    CLASS 11-D
    PRESENT

    ReplyDelete
  10. Good morning ma'am.. I have done the test and submitted attendance too! SHAMBHAVA, 11D

    ReplyDelete
  11. Good morning ma'am
    Yashas Yadav 11-D
    Present

    ReplyDelete
  12. Good morning ma'am
    Samarth Jain
    11 D
    Present

    ReplyDelete
  13. good morning
    Bhumik Tandon of class 11th D is present

    ReplyDelete
  14. Good morning ma'am
    Anugrah Singh 11D
    Present

    ReplyDelete

Post a Comment

Popular posts from this blog

NUMERICAL PRACTICE

DAY 5 27.4.20 Good Morning Boys, Today we will be practising the numericals related to the concept learnt in previous classes. Learning outcome : students will be able to Apply the formula interpret the question. Analyse the question. Molarity formula with  Density  is given as conversion table for reference : 1 cm 3 1mL 1000 mm 3 1 Litre 1000mL 1000 cm 3 1 m 3 10 6   cm 3 1000 L 1 dm 3 1000 cm 3 1 L NUMERICAL QUESTIONS : Question 1 Calculate the molarity of  N a O H N a O H  in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. Solution  Molarity M = Moles of solute(g)/ volume of solution (litres) = (4 g / 40 g ) ×  (1000/250) =0.1 mol/0.250 L = 0.4 mol/L = 0.4 M Question 2 20% (w/W) NaOH solution (density of solution= 1.2 g/cc). Calculate the Molarity? Solution 20% (w/M) means 20 gm  in 100gm  of solution Now Density of solution =1.2g/cc Density = mass /...

Chemistry Practical 11 D

Day 4 18.4.20 Good Morning students:  Today is our first chemistry practical period through blogs. please read the general instruction carefully. General Instructions :  1. students must get the concern slip signed by the parents for working in the laboratory. The concern slip contains the guidelines for working in the laboratary. Written guidelines for working in chemistry lab would be given once the school reopen. kindly note signature of  parent on concern slip is mandatory for working in chemistry lab. 2. Labcoat is mandatory for working in the lab. 3. carry your practical file and stationary during the practical period. 4. Follow the Instruction given by teacher at all times. 5. Casual attitude while working in lab will not be entertained. 6. Class would be divided into 2 batches: Batch 1 Physics Batch 2 chemistry (depending on the strength of class ) At present both the batches would be writing the practical in your chemistry file. The p...

NOMENCLATURE OF ELEMENTS

15.5.20 DAY 5 Good Morning  Boys, Today we will learn the rules for naming elements with atomic number greater than 100. Learning Outcome: student will be able to Learn the rule Apply Rule to element with Atomic number greater than 100 1.A systematic Nomenclature is derived directly from the atomic number of the elements using the numerical roots for 0 to 9. 2.The Root word are put together inorder of digits which make up the atomic number. 3. ium is added as a suffix. The Table Below will enchance your learning on nomenclature of elements of a given atomic number. THATS ALL FOR THE DAY MARK YOUR ATTENDANCE ON THE FORM Loading…