The number of atoms in 4.25 g of NH3 is approximately
(D) Weight of NH3 = 4.25g
Number of moles of NH3 = =0.25 mol
Number of molecules in 0.25 mole of NH3 = 0.25 x 6.023 x 1023
So, number of atoms = 4 x 0.25 x 6.023 x 1023
=6.0 x 1023
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The number of atoms in 4.25 g of NH3 is approximately
(D) Weight of NH3 = 4.25g
Number of moles of NH3 = =0.25 mol
Number of molecules in 0.25 mole of NH3 = 0.25 x 6.023 x 1023
So, number of atoms = 4 x 0.25 x 6.023 x 1023
=6.0 x 1023
In the reaction, 4NH3(g)+ 5O2(g) 4NO(g) +6H2O(l)
When 1 mole of ammonia and 1 mole of O2 are made to react to completion, then
(c) 4NH3(g)+ 5O2(g) 4NO(g) +6H2O(l)
4mol 5 mol 4mol 6mol
According to the equation,
5 moles of O2 required = 4 moles of NH3
1 mole of O2 required =4/5= 0.8 moles of NH3
While 1mole of NH3 requires = 5/4=1.25 moles of O2
As there is 1 mole of NH3 and 1 mole of O2 , so all the oxygen will be consumed.
An element, X has the following isotopic composition:
200X : 90%, 199X : 8.0%, 202X : 2.0%
The weighted average atomic mass of the naturally occurring element X is closest to
(d) Weight of 200X = 0.90 x 200 = 180.00 u
Weight of 199X =0.08 x 199=15.92 u
Weight of 202X = 0.02 x 202=4.04 u
Total weight = 199.96 = 200u
The hydrated salt Na2SO4.nH2O, undergoes 55% loss in mass on heating and becomes anhydrous. The value of n will be:
Na2SO4.10H2O
12 g of Mg (atomic mass=24) will react completely with an acid to give:
(b) Mg + 2HCl MgCl2 + H2
Therefore,
12g Mg gives 1/2 mole of H2
The molality of 15% (w/vol.) solution of H2SO4 of density 1.1g/cm3 is approximately:
(d) m = =1.6
H3PO4 is a tribasic acid and one of its salts is NaH2PO4. What volume of 1 M NaOH should be added to 12g NaH2PO4 (Molar mass 120) to exactly convert it into Na3PO4?
(c) Meq. of NaH2PO4 =Meq. of NaOH;
Thus, 12/120/2*1000 = 1 x V
V = 200 mL
Which mode of expressing concentration is independent of temperature?
(d) The terms which involves only masses in their formula
[e.g., molality = mass of solute x 1000 / molar mass of solute x mass of solvent]
are independent of temperature. On the other hand, since, volume changes with temperature, the terms having volume in their formula [e.g., molarity = mass of solute x 1000 / molar mass of solute x volume of solvent],NORMALITY are dependent on temperatue.
The product of atomic mass and specific heat of a metal is approximately 6.4. This was given by:
(d) Dulong Petit's law: atomic mass x sp. heat =6.4
0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is:
(c) Meq. of H2SO4 + Meq.Of SO3 = Meq. of NaOH
(0.5-a)/49*1000 +a/40*1000 = 26.7 x 0.4
a = 0.103
% of SO3 = 0.103/0.5*100 = 20.6%
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