Molar Mass of H2O = 2 + 16 = 18 g/moles
So ,number of moles of H2O = Mass/Molar Mass = 54/18 =3 moles
Now 1 moles = 6.022×1023 molecule
So 3 moles will have 18.066×1023 molecule
Molar mass (Molecular mass in gram) of NH4Cl= 14+4+35.5= 53.5g
No. of moles of NH4Cl
=6.022×10236.022×1023
= 1 mole
Mass of NH4Cl
=No. of moles×molar mass
=1×53.5g=53.5g.
No. of moles of Oxygen atoms
=12.044×10236.022×1023
= 2 mole
Mass of Oxygen atoms
=No. of moles×atomic mass
=2×16
= 32 g
Molar mass (Molecular mass in gram) of NH4= 14+4= 18 g
No. of moles of NH4?
= 36/18 = 4 moles
Now Total Moles of Hydrogen Atoms
= 16 moles
=16x6.022×1023=96.352×1023
1 molecule of H2= 2 hydrogen atoms
So, 1 mole of H2= 2 mole hydrogen atoms
=2×6.022×1023=12.044×1023 hydrogen atoms.
No. of moles of Cu
=Mass of CuAtomic mass
= 0.3175/63.5
=0.005 mole
No. of Cu atoms
=No. of moles×Avogadro constant
=0.005×6.022×1023
=30.11×1020 Cu atoms.
1 mole of CH4= 22.4 L (at NTP)
Therefore
22.4 L of CH4gas= 1 mole CH4gas= 6.022×1023 CH4 gas molecules.
a. Molar Mass of H2gas = 2 gm
Number of moles=MassMolar Mass=12=.5
1 mole of H2= 22.4 L (at NTP)
Therefore
.5 mole of H2= .5 x 22.4 = 11.2 litre
b.
No. of moles of H2= 20/2 =10
1 mole of H2= 22.4 L (at NTP)
Therefore
10 moles = 10×22.4
=224 L
c.
6.022×1023 molecules = 1 mole molecules, and
1 mole molecules of any ideal gas occupies 22.4 L at NTP.
Mass of 1 mole Y atoms
=mass of 1 atom×Avogadro constant
=6.644×10−23×6.022×1023
= 40 g
So, the atomic mass of Y = 40
No. of gram-atoms (or moles) of X
=mass of Yatomic mass
=40×100040=1000
1 mole of H2S contains 2 mole of H, 1 mole of S
Therefore
10 mole of H2S contains
20 mole of H = 20×6.022×1023=12.044×1024 H atoms
10 mole of S = 10×6.022×1023=6.022×1024 S atoms
Molecular mass of KClO3= 39+35.5+3×16=122.5
No. of mole of KClO3= 245 g/122.5g = 2 mole
2 mole of KClO3contains
2 mole of K = 2×6.022×1023=12.044×1023 K atoms
2 mole of Cl = 2×6.022×1023=12.044×1023 Cl atoms
6 mole of O = 6×6.022×1023=1.806×1024 O atoms.
Mass of NaOH = 4 g
Number of moles of = 4 g NaOH/40 g = 0.1 mol
Mass of H2O = 36 g
Number of moles of = 36 g/18= 2 mol
Mole fraction of water=Number of moles of water No. of moles of water+No. of moles of NaOH
=22+.1=0.95
Mole fraction of NaOH=Number of moles of NaOHNo. of moles of NaOH+ No. of moles of water
=0.12+0.1=0.047
Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g
Volume of solution = 40×1=40 mL (Since specific gravity of solution is = 1 g m/L)
Molarity of solution=Number of moles of solute Volume of solution in litre=0.10.04=2.5 M
No. of moles in 12.2 L gas at NTP = 12.2/22.4 = 0.50
No. of molecules in 12.2 L gas
=0.50×6.022×1023
=3.011×1023 molecules
In mono-atomic gases, No. of atoms = No. of molecules =3.011×1023.
In diatomic gases, No. of atoms = 2×No. of molecules=2×3.011×1023=6.022×1023.
Mass of one mole of oxygen molecule (O2)
= molecular mass of oxygen molecule (O2) in gram
= 32 g.
Mass of a single oxygen molecule
=321×1024
=3.2×10−23 g
Standard molar volume is Volume of 1 mole of gas i.e 32 gm O2 gas
Now
Density=Massvolume
Volume=MassDensity
= 32/1.429
=22.39 Litres.
Molar mass pf methane =16
Number of Moles =25/16
No of methane molecules = 2516×6.022×1023=9.411×1023
No of carbon molecules = 1×9.411×1023=9.411×1023
No of hydrogen molecules = 4×9.411×1023=3.74×1024
Mass of 1 mole metal atoms = 54.94 g
Mass of 1 metal atom = 54.946.022×1023=9.12×1023g
Volume occupied by one metal atom
=Mass of one metal atomdensity
=9.12×10237.42
=1.23×1023 cc.
Molecular mass of CaCO3= 40+12+3 \times 16 = 100
No. mole of CaCO3= 200g100g=2
Number of Moles of Ca=2 moles
Number of Moles of C=2 moles
Number of Moles of O=6 moles
Molecular mass of CH4= 12+4×1=16
Moles of CH4=3.216=0.2
No. of electron in 1 molecule of CH4= 6+4 = 10 electrons
Total no. of electrons = 0.2×6.022×1023×10=12.044×1023 electrons.
Number of Moles of Nitrogen = 2014=1.428 moles
From the Molecular formula, it is apparent that, the number of Moles Ca will be half number of Number of Nitrogen
So, number of Moles of Ca=1.4282=.714 moles
So mass of Ca = .714×40=28.56 g
1 gm atom = atomic weight
So number of atoms will be Avogadro number = 6.022×1023
The molar ratios are also volume ratios for gases (Avogadro’s principle)
Molecular mass of air
=78×28+21×32+0.9×40+0.1×4478+21+0.9+0.1
=28.964.
Molar Mass of Ammonia= 14+3×1=17 gm
Mass of One molecule =17(6.022×1023)=2.8×10−23 g
No. of mole of P = x20
No. of atoms of P = (x20)×N [N is Avogadro constant]
Therefore, y=x×N20
or x=20yN
Now,
No. of mole of Q = 2x40
No. of atoms of Q
=(2x40)×N
=2N40×20yN
=y
Pressure = 7.6×10−10 mm Hg
as 1 atm = 760 mm Hg
p=7.6×10−10760
=10−12 atm
Volume = 1 litre
Temperature = 0°C = 273 K
We know pV=nRT
or
n=pVRT
n=(10−12×1)(0.0821×273)=0.44×10−13
No. of molecules=no. of moles×Avogadro constant
=0.44×10−13×6.022×1023
=2.65×1010 .
Volume ratio
= Molar ratio (Avogadro’s principle – the molar ratios are also volume ratios for gases)
=1:1
Molar mass (i.e., molecular mass in g) = 40+12+3×16=100g
Mass of .5 moles of CaCO3= .5×100=50 g
Molar Mass of NaCl= 23+35.5=58.5 g/mol
Cost of 1 Mole of NaCl =(101000)×58.5=.585 Rs/mole
Molar Mass of C12H22O11= 12×12+22×1+11×16=342 g/mol
Cost of 1 Mole of C12H22O11 = (401000)×342=13.68 Rs/mole
Moles of oxygen atoms in 1 mole of Na2CO3.10H2O = 3+10 = 13
Moles of oxygen atoms in 0.2 mole of Na2CO3.10H2O = 0.2×13=2.6
Therefore, Number of oxygen atoms = 2.6×6.022×1023=1.565×1024.
Let the mass of polystyrene be 100g,
Therefore, No. of moles of Br in 100g of polystyrene = 10.46/79.9 = 0.1309
From the formula Br3C6H3(C3H8)n, we have,
No. of moles of Br=3×moles ofBr3C6H3(C3H8)n
0.1309=3×massmolecular mass=3×100314.5+44n
n = 44.9
n=45(approx)
It is given that out of 100 atoms, 93 atoms are Hydrogen and 7 atoms are Helium.
Mass of Hydrogen atoms = 93×1=93
Mass of Helium atoms = 7×4=28
Therefore, Mass percentage of Hydrogen = 9393+28×100=76.86%
Mass percentage of Helium = 2893+28×100=23.14%.
Mass of Fe in haemoglobin
= 0.35% of 65000
=0.35×65000100
= 227.5
Therefore, No. of Fe atoms in a haemoglobin molecule = 227.556=4
Moles of CO2 in 88 g = 88/44 = 2
Number of Oxygen atom=2×6.022×1023×2=20.092×1023
Now
Number of molecules of CO having 20.092×1023 of oxygen atom will be 20.092×1023
Hence of weight of CO = 28×20.092×10236.022×1023|=112g
Let the volume of ethanol containing the same number of molecules as are present in 175mL of water be V mL.
Moles of C2H5OH in V mL = Moles of H2O in 175 mL
Mass of C2H5OH /mol. mass of C2H5OH = Mass of H2O / mol. mass of H2O
Now as Density=Massvolume or Mass=Density×Volume
Therefore
0.789×V46=1×17518
V =566.82 mL.
Mass % of Mg = 2.68%
Now 100 g of Chlorophyll contains 2.68 g of Mg
Then 4 g of Chlorophyll contains = (2.68100)×4=.1072 g
Now
Number of atoms of Mg =MassMolar mass×6.022×1023=1.34×1021 atoms
Suppose the side of cube = x cm = diameter of mercury atom
Therefore, Volume of 1 Hg atom = x3 and
Mass of 1 Hg atom
=density×volume
=13.6×x3
Mass of 1 Hg atom
= Atomic mass/ Avogadro constant
=2006.022×1023
13.6×x3=2006.022×1023
x=(2.44×10−23)1/3=2.9×10−8 cm.
Rupees= 6.022×1023
Rupees spend rate = 106 /sec
Rupees spent per year =106×3600×24×365
Number of Years required to fully spend the money
=6.022×1023106×3600×24×365
=1.90988×1010 years
1 mole of 14C=14 g
No. moles of carbon atoms = 7 mg/14 g = 5×10−4 Moles
Number of atoms of carbon atoms = 6.022×1023×5×10−4=3.01×1020
A C-14 atom contains 8 neutrons
Therefore, The total number of neutrons = 8×3.01×1020=2.408×1021.