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10L of hard water required 5.6g of lime ...

10L of hard water required 5.6g of lime for removing hardness. Hence temperorary hardness in ppm of `CaCO_(3)` is:

A

1000

B

2000

C

100

D

1

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • 10 L of hard water required 0.56 g of lime (CaO) for removing hardness. Hence, temporary hardness in ppm (part per million 10^(6) ) of CaCO_(3) is:

    A
    100
    B
    200
    C
    10
    D
    20

  • 10 L of hard water requried 0.56 g of lime (CaO) for removing hardness. Hence, temporary hardness in ppm (part per million, 10^(6) ) of CaCO_(3) is

    A
    100
    B
    200
    C
    10
    D
    20

  • 10L of hard water required 0.56g of time (CaO) for removing hardness. Hence, temporary hardness in p p m (part per million, 10^6 ) of CaCO_3 is

    A
    `100`
    B
    `200`
    C
    `10`
    D
    `20`

    • Similar Questions

    Explore conceptually related problems

    One kilogram sample of hard water contains 4.44 mg of CaCl_(2) and 1.9 mg of NaCl. The total hardness in tems of ppm of CaCO_(3) is :

    How is temporary hardness in water removed?

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