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Question: In one process for waterproofing, a fabric is exposed to \({(C{H_3})_2}SiC{l_2}\) vapour. The vapour...

In one process for waterproofing, a fabric is exposed to (CH3)2SiCl2{(C{H_3})_2}SiC{l_2} vapour. The vapour reacts with hydroxyl group hydroxyl groups on the surface of the fabric or with traces of water to form the waterproofing film [(CH3)2SiO]n{[{(C{H_3})_2}SiO]_n}, by the reaction n(CH3)2SiCl2+2nOHnH2O+[(CH3)2SiO]3n{{{(C}}{{{H}}_{{3}}}{{)}}_{{2}}}{{SiC}}{{{l}}_{{2}}} + 2nO{H^ - } \to n{H_2}O + {[{(C{H_3})_2}SiO]_3} where n stands for a large integer. The waterproofing film is deposited on the fabric layer upon layer. Each layer is 6.0A06.0{A^0} thick [(CH3)2SiO]n{[{(C{H_3})_2}SiO]_n} , the thickness of [(CH3)2SiO][{(C{H_3})_2}SiO] group. How much (CH3)2SiCl2{(C{H_3})_2}SiC{l_2} is needed to waterproof one side of a piece of fabric, 1.00m1.00m by 3.70m3.70m, with 300 layers thick? The density of the film is 1.0g/cm31.0g/c{m^3}. ( Si=28Si = 28)
A. 3.1×10193.1 \times {10^{19}} units
B. 4.1×10194.1 \times {10^{19}} units
C. 2.6×10272.6 \times {10^{27}} units
D. 4.5×10224.5 \times {10^{22}} units

Explanation

Solution

We know the relation between density, mass, and volume Density=MassVolumeDensity = \dfrac{{Mass}}{{Volume}}. Another formula used is to find the number of moles, Number  of  moles=given  massmolar  mass=wbmbNumber\;of\;moles = \dfrac{{given\;mass}}{{molar\;mass}} = \dfrac{{{w_b}}}{{{m_b}}}. We know that
Number of molecules or number of atoms =No.of  moles(n)×Avogadro  number(NA) = No.of\; moles(n) \times Avogadro\;number({N_A})

Complete step by step answer:
Here, the dimensions of one piece of fabric are given. It is 6.0A06.0{A^0} thick, 1.00m1.00mby 3.70m3.70m
So, the volume can be calculated.
6.0A0=6×1010cm6.0{A^0} = 6 \times {10^{ - 10}}cm
Here, we have converted meters into centimeters, 1.00m=100cm1.00m = 100cm and 3.70m=370cm3.70m = 370cm
Thus, the volume of one film can be found out as l×b×hl \times b \times h
The volume of 1 file is, volume =(100×370×6×1010)cm = (100 \times 370 \times 6 \times {10^{ - 10}})cm
Volume of one film =222000×1010 = 222000 \times {10^{ - 10}}
volume=2.22×105cm3\Rightarrow volume = 2.22 \times {10^{ - 5}}c{m^3}
Here, the density of the film is given 1.0g/cm31.0g/c{m^3}. We already know the relationship between density, mass, and volume, so the mass of 1 film can be found out.
With the help of volume and density, the value of Mass of 1 film is 2.22×105g2.22 \times {10^{ - 5}}g
There are films with 300 layers thick, so the mass of 300 layers can be found by multiplying 300 with a mass of 1 film.
So, the mass of 300 films can be written as, mass=300×2.22×105g=6.66×103gmass = 300 \times 2.22 \times {10^{ - 5}}g = 6.66 \times {10^{ - 3}}g
We know that Number of molecules or number of atoms =n×NA = n \times {N_A}
We need to find the amount of (CH3)2SiCl2{(C{H_3})_2}SiC{l_2} required to waterproof one side of the fabric.
The molar mass (CH3)2SiCl2{(C{H_3})_2}SiC{l_2} is 129 g/mol.
Hence, the number of moles of (CH3)2SiCl2{(C{H_3})_2}SiC{l_2}is, Number  Of  Moles=wbmbNumber\; Of\; Moles = \dfrac{{{w_b}}}{{{m_b}}}
Substituting the value of given mass and molar mass of the molecule we have
No. of moles of (CH3)2SiCl2{(C{H_3})_2}SiC{l_2} =6.66×103129 = \dfrac{{6.66 \times {{10}^{ - 3}}}}{{129}}
Number of (CH3)2SiCl2{(C{H_3})_2}SiC{l_2} required to waterproof one side of a piece of fabric=6.66×103129×6.022×1023 = \dfrac{{6.66 \times {{10}^{ - 3}}}}{{129}} \times 6.022 \times {10^{23}}
Number of (CH3)2SiCl2{(C{H_3})_2}SiC{l_2} =3.1×1019 = 3.1 \times {10^{19}}units

So, the correct answer is Option A.

Note: Waterproofing is a method of making a substance resistant to water. Waterproof substances are unaffected by water. The application includes use in buildings, waterproof paints, parts of the ship, clothes, etc. Recent studies have shown that some nanotextured surfaces that use cone forms can produce highly water repellent surfaces which are highly hydrophobic.