45) using ideal gas equation , PV= nRT V=25L , P=702mmHg = 702/760 atm, T=22.5 o
ID: 700361 • Letter: 4
Question
45) using ideal gas equation , PV= nRT
V=25L , P=702mmHg = 702/760 atm, T=22.5oC = 295.5K
R, gas constant =0.0821Latm/mol/K
n =( 702 *25)/ (760*0.0821*295.5) = 0.95mol
46) According to stoichiometry of given reaction,
149g of KCl is obtained by 111 g CaCl2
So 10g KCl is obtained by 111*10/149 = 7.44g of CaCl2
47) P=780/760atm, T=135+273 = 408K, V=0.5L
n= PV/RT = (780*0.5)/(760*408*0.082) = 0.015mol
No.of molecules= no.of moles* 6.022*1023
=0.015*6.022 *1023 = 9.23 * 1021
48) P1 T2 = P2 T1
1* (475+273) =P2 * 298
P2 = 2.51atm
49) Accor. To stoichiometry of reaction,
1mol N2. Requires 3mol H2
so 6mol N2 requires 18mol H2
but we have only 6mol H2 so H2 is limiting reagent.
3mol H2 forms 2mol NH3
So 6 mol H2 forms 4mol NH3.
vol. Of 1mol NH3 at STP = 22.4L
SO VOL. of 4mol NH3 = 22.4* 4 = 89.6L
50) hydrogen has highest first ionization enthalpy as its e- is present in first shell feels high nuclear charge thus large amount of energy is required to remove it.