In one process for waterproofing, a fabric is exposed to (CH3)2SiC12 vapor. The

Question

In one process for waterproofing, a fabric is exposed to (CH3)2SiC12 vapor. The vapor reacts with hydroxyl groups on the surface of the fabric or with traces of water to form the waterproofing film [CH3)2SiO]n by the reaction n (CH3)2SiC12 + 2n OH-1 rightarrow 2n Cl-1 + n H2O + [(CH3)2SiO]n where n stands for a large integer. The waterproofing film is deposited on the fabric layer by layer. Each layer is 0.6 nm think the thickness of the (CH3)2SiCl2 group). How many grams of (CH3)2SiC12 is needed to waterproof a piece of fabric 1.00 m by 3.00 m with a film thickness of 300 layers. The density of the film is 1.0 g/cm3.

Explanation / Answer

First, we have to calculate the area of the fabric that is to be coated.

Area= length*breadth

area= 3 m*1 m= 300cm*100cm= 30,000 cm2

Now, Thickness of one layer of the waterproofing film , (CH3)2SiCl2 group,

Given is 0.6nm= 6*10-8 cm

Now no. of layers needed= 300

so, total thickness= 300 layers

i.e. thickness of the fabric after waterproofing= 30,000*6*10-8*300 cm3= 0.54 cm3

density of film= 1g/cm3.

Now, we know that,

Density= mass/volume

so, mass= density*volume= 1*0.54 g= 0.54 g

So, 0.54 grams of (CH3)2SiCl2 is needed.

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