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mg/l to ppm

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  • mg/l to ppm

    HI, can someone tell me the conversion to convert mg/l into parts per million (ppm)? my first challenge is converting 9.52 mg/l into ppm

    thanks,

    Craig

  • #2
    Re: mg/l to ppm

    Originally posted by Unregistered View Post
    HI, can someone tell me the conversion to convert mg/l into parts per million (ppm)? my first challenge is converting 9.52 mg/l into ppm

    thanks,

    Craig
    Assuming a dilute solution in water, 1 L weighs 1 kg, and 1 ppm = 1 mg/L, This is also true if ppm is computed on a weight/volume basis.

    If ppm is weight/weigh and is not a water solution or no dilute, it needs to be corrected by the density of the mixture.

    Comment


    • #3
      Re: mg/l to ppm

      hello John, and thanks for getting back to me. In reference to a soil analysis, am I to understand that 9.37 mg/l is equal to 9.37 ppm?

      Craig

      Comment


      • #4
        Re: mg/l to ppm

        Originally posted by Unregistered View Post
        hello John, and thanks for getting back to me. In reference to a soil analysis, am I to understand that 9.37 mg/l is equal to 9.37 ppm?

        Craig
        Not necessarily. My remarks were true for solutions.

        However, soil analysis usually involves a measured amount of soil mixed in a measured amount of water. The dissolvable nutrients dissolve in the water. The concentration of nutrients in the water is analyzed as described, however, it is usually desirable to determine the concentration in the soil, not what develops in the water.

        The concentration in water has to be multiplied by the total volume of water used in the leeching to get total mass of the substance in question. That mass has to be divided by the total soil mass.

        Let us suppose you got the 9.37 mg/L as a result of mixing 10 g of soil in 100 mL (0.1 L) of water, then:
        1) 9.37 mg/L x 0.1 L = 0.937 mg
        2) 0.937 mg/ 10 g = 93.7 g/g = 93.7 ppm (weight/weight)

        You need to know the weight of soil sample and the volume of water used to leech it as well as the concentrations.

        Comment

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