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A particle of suspension of radius 1 mm is broken to form colloidal particles of radius 1000 Å. How many times will be the total surface the colloidal particles as sompared to the surface area of the particle of suspension. |
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Answer» Solution :Volume of paticle of suspension of RADIUS 1 mmm `=4/3pi r^(3)=4/3pi (0.1)^(3)cm^(3)=4/3pi xx10^(-3)cm^(3)` Volume of particle of radius `1000Å (10^(-5)cm)=4/3pi (10^(-5))^(3)=4/3pixx10^(-15)cm^(3)` `therefore` NUMBER of colloidal particles formed `=((4//3)pi xx10^(-3))/((4//3)pi xx10^(-15))=10^(12)` Surface area of each colloidal particle `=4 pi r^(2)=4 pi (10^(-5))^(2) cm^(2)=4 pi xx10^(-10)cm^(2)` `therefore` Total surface area of `10^(12)` particles `=(4pi xx10^(-10))xx10^(12)= 4pi xx10^(2)cm^(2)` Surface area of particle of suspension `=4 pi r^(2)=4 pi (0.1)^(2)cm^(2)=4 pi xx10^(-2)cm^(2)` `therefore ("Total surface area of colloidal particles")/("Surface area of particle of suspension")=( 4pi xx10^(2))/(4 pi xx10^(-2))=10^(4)` |
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