(4) 50 grams of dry soil are suspended in 1 liter of dispersing solution. The su
ID: 542432 • Letter: #
Question
(4) 50 grams of dry soil are suspended in 1 liter of dispersing solution. The suspension is agitated, and allowed to settle. A hydrometer reading of 38 is obtained. The suspension is re-agitated and allowed to settle for 2 hours, a hydrometer reading of 22 is obtained. The "blank" reading (for the dispersing solution in the absence of soil) is 5. The hydrometer reads out in units of grams per liter of suspended matter. a. How many grams of (silt + clay) were in this sample? b. How many grams of sand were in this sample? o. What was the percent stand in this sample? d. How many grams of clay were in this sample? e. What was the percent clay in this sample? £ what was the percent silt in this sample? 8. According to the USDA soil textural triangle (available online or in your text, or in your lab handout), what was the texture of this soil?Explanation / Answer
Hydrometer analysi of the given soil sample gives an estimation about the content of the soil basis the size of the particles (using Stoke's law).The velocity of settlement of the grains in the sample soil depends on the weight ,size and shape ,with the coarser one settling faster.
Thus,firstly sand particles settles out of suspension ,next goes silt and then clay(finest particles)
a)Blank reading of hydrometerfor the dispersing solution =5g/L
Reading after first agitation=38g/L
Thus,mass of silt+clay =38 g
b) massof sand =mass of dry soil- mass of silt+clay =50-38=12 g
c)percent sand=(mass of sand/total mass of sample)*100=(12/50)*100=24%
d) mass remaining in the suspension after second agitation (clay)=22 g
mass of clay =22g
e)percent clay=(22/50)*100=44%
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