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10 L of hard water required 0.56 g of li...

`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

Text Solution

Verified by Experts

The correct Answer is:
B
Temporary hardness is due to `HCO_(3)^(-)" of "Ca^(2+) and Mg^(2+)`
`Ca(HCO_(3))_(2)+underset(56g)(CaO)rarr underset((2xx100g))(2CaCO_(3))+H_(2)O`
`0.56g CaO-=2g CaCO_(3)" in "10 L H_(2)O`
`=2g CaCO_(3)" in "10^(4)mL H_(2)O`
`=200g CaCO_(3)" in "10^(6)mL H_(2)O`
`=200" ppm of "CaCO_(3)`
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Knowledge Check

  • 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`

  • 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

  • Temporary hardness in water is removed by

    A
    boiling
    B
    filtration
    C
    distillation
    D
    decantation

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