A vessel contains two non-reactive gases, neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3.2. Estimate the ration of (i) number of molecules and (ii) mass density of neon and oxyge in the vessel. Atomic mass of Ne- 20.2 u, molecular mass of `O_2` = 32.0 u.

Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is u_rms the largest ?

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 ^@C . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 ^@C . Estimate the mass of oxygen taken out of the cylinder ( R= 8.31 J mol^-1 K^-1 ’, molecular mass of O_2 = 32 u)

How many grams of oxygen gas contain the same number of molecules as are present in 16 grams of sulphur dioxide ? (Atomic masses : S = 32, O = 16).

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 ^@C . Take the radius of a nitrogen molecule to be roughly 1.0overset@ A . Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N_2 = 28.0 u).

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 For the reaction, Fe_0.95O("molar mass=M") to Fe_(2)O_(3) what is the eq. mass of fe_(0.95) O ?

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 Consider the following reaction. H_(3)PO_(2)+NaOH to NaH_(2)PO_(2)+H_(2)O What is the equivalent mass of H_(3)PO_(2) ?(mol.Wt.is M)

The vapour pressure of pure bronze at a certain temparature is 262atm. At the same temparature the V.P. of a solution containing 2.0g of non-volatile solid in 100g bronze is 256atm . What is the molecular mass of the solid?

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 In the reaction, xVO+yFe_(2)O_(3) to FeO+V_(2)O_(5) what is the value of x and y respectively?

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predict the molar ratio of the reactant species taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 n-factor of Ba(MNO_(4))_(2) in acidic medium is :

STATEMENT-1: Gram molecular mass of O_(2) is 32. STATEMENT-2: Relative atomic mass of oxygen is 32 a.m.u.