Unformatted text preview: The total of the masses retained is 278g, which together with the 121g collected in the pan gives 399g. Thus there is a shortfall of 12g, which is attributable to sieve losses. Take the total mass of the sample as 399g. The % by mass of the sample passing each sieve is given by the total sample mass (399g) minus the cumulative mass of soil retained on larger size sieves. Hence Sieve size, mm 6.3 1.2 0.3 0.063 Mass retained, g 0 60 126 92 Cumulative mass retained, g 0 60 186 278 Mass passing, g 399 339 213 121 % passing 100 85.0 53.4 30.3 The sedimentation test data are again already partprocessed, with the mass of soil in each size range expressed as a percentage of the 121g collected in the pan. The fraction of the pan sample smaller than a given size is equal to the sum of the percentages in this and the smaller...
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 Fall '11
 StuartChalk
 Analytical Chemistry, Mass, Particle size distribution, Sieve analysis, Sedimentation Test, sieve size

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