Mole Concepts Problems with Solutions

Molar Mass of Ammonia= $14 + 3 \times 1= 17$ gm
Mass of One molecule =$ \frac { 17}{(6.022 \times 10^{23})} = 2.8 \times 10^{-23}$ g

No. of mole of P = $\frac {x}{20}$
No. of atoms of P = $(\frac {x}{20}) \times N$ [N is Avogadro constant]
Therefore, $y = x \times \frac {N}{20}$ or $x= \frac {20y}{N}$
Now,
No. of mole of Q = $\frac {2x}{40}$
No. of atoms of Q
$= (\frac {2x}{40}) \times N$
$= \frac {2N}{40} \times \frac {20y}{N}$
=y

Pressure = $7.6 \times 10^{-10}$ mm Hg
as 1 atm = 760 mm Hg
$p= \frac {7.6 \times 10^{-10}}{ 760}$
$= 10^{-12}$ atm
Volume = 1 litre
Temperature = 0°C = 273 K
We know $pV = nRT$
or
$n = \frac {pV}{RT}$
$n = \frac {(10^{-12} \times 1)}{(0.0821 \times 273)} = 0.44 \times 10^{-13}$
$\text {No. of molecules} = \text {no. of moles} \times \text {Avogadro constant}$
$= 0.44 \times 10^{-13} \times 6.022 \times 10^{23}$
$=2.65 \times 10^{10}$ .

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) = $40+12+3 \times 16 = 100$g
Mass of .5 moles of CaCO₃= $.5 \times 100 = 50$ g

Molar Mass of NaCl= $23 + 35.5 = 58.5$ g/mol
Cost of 1 Mole of NaCl =$ (\frac {10}{1000} ) \times 58.5 = .585$ Rs/mole
Molar Mass of $C_{12}H_{22}O_{11}$= $12 \times 12 + 22 \times 1 + 11 \times 16 = 342$ g/mol
Cost of 1 Mole of $C_{12}H_{22}O_{11}$ = $( \frac {40}{1000} ) \times 342 = 13.68$ 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 = $0.2 \times 13= 2.6$
Therefore, Number of oxygen atoms = $2.6 \times 6.022 \times 10^{23}=1.565 \times 10^{24}$.

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 $Br_3C_6H_3(C_3H_8)_n$, we have,
$\text{No. of moles of Br} = 3 \times \text {moles of} \; Br_3C_6H_3(C_3H_8)_n$
$0.1309 = 3 \times \frac {mass}{\text{molecular mass}} = 3 \times \frac {100}{314.5+44n}$
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 = $93 \times 1 = 93$
Mass of Helium atoms = $7 \times 4 =28$
Therefore, Mass percentage of Hydrogen = $ \frac {93}{93+28} \times 100 =76.86$%
Mass percentage of Helium = $\frac {28}{93+28} \times 100 =23.14$%.

Mass of Fe in haemoglobin
= 0.35% of 65000
$= \frac {0.35 \times 65000}{100}$
= 227.5
Therefore, No. of Fe atoms in a haemoglobin molecule = $ \frac {227.5}{56} =4$

Moles of $CO_2$ in 88 g = 88/44 = 2
Number of Oxygen atom=$ 2 \times 6.022 \times 10^{23} \times 2 = 20.092 \times 10^{23}$
Now
Number of molecules of CO having $20.092 \times 10^{23}$ of oxygen atom will be $20.092 \times 10^{23}$
Hence of weight of CO = $ \frac { 28 \times 20.092 \times 10^{23}}{6.022 \times 10^{23}}| =112$g

Let the volume of ethanol containing the same number of molecules as are present in 175mL of water be V mL.
Moles of $C_2H_5OH$ in V mL = Moles of $H_2O$ in 175 mL
Mass of $C_2H_5OH$ /mol. mass of $C_2H_5OH$ = Mass of $H_2O$ / mol. mass of $H_2O$
Now as $Density= \frac {Mass}{volume}$ or $Mass = Density \times Volume$
Therefore
$ \frac {0.789 \times V}{46} = \frac {1 \times 175}{18}$
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 = $(\frac {2.68}{100}) \times 4 = .1072$ g
Now
Number of atoms of Mg =$ \frac {Mass}{\text {Molar mass}} \times 6.022 \times 10^{23} = 1.34 \times 10^{21}$ atoms

Suppose the side of cube = x cm = diameter of mercury atom
Therefore, Volume of 1 Hg atom = $x^3$ and
Mass of 1 Hg atom
$= density \times volume$
$= 13.6 \times x^3$
Mass of 1 Hg atom
= Atomic mass/ Avogadro constant
$= \frac {200}{6.022 \times 10^{23}}$
$13.6 \times x^3= \frac {200}{6.022 \times 10^{23}}$
$x = (2.44 \times 10^{-23})^{1/3}=2.9 \times 10^{-8}$ cm.

Rupees= $6.022 \times 10^{23}$
Rupees spend rate = 10⁶ /sec
Rupees spent per year =$10^6 \times 3600 \times 24 \times 365$
Number of Years required to fully spend the money
$=\frac {6.022 \times 10^{23}}{ 10^6 \times 3600 \times 24 \times 365}$
$=1.90988 \times 10^{10}$ years

1 mole of ¹⁴C=14 g
No. moles of carbon atoms = 7 mg/14 g = $5 \times 10^{-4}$ Moles
Number of atoms of carbon atoms = $6.022 \times 10^{23} \times 5 \times 10^{-4}= 3.01 \times 10^{20}$
A C-14 atom contains 8 neutrons
Therefore, The total number of neutrons = $8 \times 3.01 \times 10^{20} =2.408 \times 10^{21}$.