CBSE ADDA

chemistry adda: 9th chemistry Self-evaluation on Mole concept Sample Problem 1.  Calculate the number of mole in 52 g of Helium. Solution.  We know that, Atomic mass of He = 4u So, its molar mass = 4 g  That is, 4 g of He contains 1 mole of He. Or, 4 g of He = 1 mole of He Or, 1 g of He = 1/4 mole of He So, 52 grams of Helium =1/4 x 52 mole = 13 moles Therefore, there are 13 moles in 52 g of He Sample Problem 2.  Calculate the number of moles for 12.044 X 10 23  atoms of Helium. Solution:  6.022 X 10 23  atoms of Helium           = 1 mole 1 atom of Helium                                 =1/6.022 X 10 23  mole 12.044 X 10 23  atoms of Helium         = [(1/6.022 X 10 23 ) x    12.044 X 10 23   ]moles [ You may use  ----        Number of moles    = (given numbers of particles /Avogadro number)] Sample Problem 3.  How many moles are there in 5 grams of calcium? Sample Problem 4.  How many moles there in 12.044 X 10 23  atoms of phosphorous? Samp

CBSE CHEMISTRY CLASS 9TH CHAPTER - 3 ATOMSAND MOLECULES Download solved paper based on mole concept 9th Mole concept numerical problems solved-CBSE SET- 1 Download File 9th Mole concept numerical problems solved-CBSE SET- 2 Download File practice your learning  1. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24g/ mol 2. Verify by calculating that (a) 5 moles of CO 2 and 5 moles of H 2 O do not have the same mass. (b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3:5. 3. Find the ratio by mass of the combining elements in the following compounds.  (a) CaCO 3 (b) MgCl 2 (c) H2SO4 (d) C2H 5 OH  (e) NH 3   (f) Ca(OH) 2 4. Calcium chloride when dissolved in water dissociates into its ions according to the following equation. CaCl 2 (aq) → Ca 2 + (aq) + 2Cl– (aq) Calculate the number of ions obtained from CaCl 2 when 222 g of it is dissolved in water.

WORK AND ENERGY   9th physics Term-II 1. When a body falls freely towards the earth, then its total energy (a) increases (b) decreases (c) remains constant (d) first increases and then decreases 2. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car (a) does not change (b) becomes twice to that of initial (c) becomes 4 times that of initial (d) becomes 16 times that of initial 3 . In case of negative work the angle between the force and displacement is (a) 00 (b) 450 (c) 900 (d ) 1800 4 . An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same (a) acceleration (b) momenta (c) potential energy (d) kinetic energy 5 . A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational forc

CBSE PHYSICS: LAW OF CONSERVATION OF ENERGY(9th physics) : LAW OF CONSERVATION OF ENERGY Energy can neither be created nor destroyed, but it is transformed from one form to another. Alternatively, whenever energy gets transformed, the total energy remains unchanged. Proof – Freely falling body It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant. Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig. At A, Kinetic energy kE = 0 Potential energy Ep = mgh Total energy E = Ep + Ek = mgh + 0= mgh During the fall, the body is at a position B. The body has moved a distance x from A. At B, velocity v 2 = u 2 + 2as applying, v 2 = 0 + 2ax = 2ax Kinetic energy Ek = 1/2 m v 2 = 1/2 m x 2gx = mgx Potential energy Ep = mg (h – x) Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh If the body reaches the position C. At C, Potential energy Ep = 0 V