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Standard XII

Chemistry

Hardness of Water and Treatment

10 L of hard ...

Question

$10$ L of hard water required $0.56$ g of lime (CaO) for removing hardness. Hence, the temporary hardness in ppm (part per million) of $C a C O_{3}$ is :

100

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200

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10

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20

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Solution

The correct option is B $200$
Temporary hardness is due to $H C O_{3}^{⊖}$ of $C a^{2 +}$ and $M g^{2 +}$ .
$C a (H C O_{3})_{2} + \frac{C a O}{56 g} \to \frac{2 C a C O_{3}}{(2 \times 100) g} + H_{2} O$
$0.56$ g $C a O \equiv 2$ g $C a C O_{3}$ in $10$ L $H_{2} O$
$= 2$ g $C a C O_{3}$ in $10^{4}$ mL $H_{2} O$
$= 200$ g $C a c O_{3}$ in $10^{6}$ mL $H_{2} O$

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Similar questions

10

L of hard water required

0.56

g of lime

(C a O)

for removing hardness. Hence, temporary hardness in ppm of

C a C O_{3}

is :

10

L of water requires

0.28

g of lime (CaO) for removing hardness. Calculate the temporary hardness in ppm of

C a C O_{3}

Q. Hardness of water is measured in terms of ppm (parts per million) of

C a C O_{3}

. It is the amount (in g) of

C a C O_{3}

present in

10^{6} g H_{2} O

. In a sample of water,

10

L required

0.56

g of

C a O

to the remove temporary hardness of

H C O_{3}^{-}

C a (H C O_{3})_{2} + C a O ⟶ 2 C a C O_{3} + H_{2} O

Temporary hardness is :

10

L of hard water with temporary hardness of

[C a (H C O_{3})_{2}]

requires

0.56

g of lime. The reaction follows as:

C a (H C O_{3})_{2} + C a O ⟶ 2 C a C O_{3} + H_{2} O

.
Temporary hardness in terms of ppm of

C a C O_{3}

is:

Following reaction is used to remove temporary hardness which is due to dissolved bicarbonates of

C a^{2 +}

and

M g^{2 +}

C a (H C O_{3})_{2} + C a (O H)_{2} \to 2 C a C O_{3} ↓ + 2 H_{2} O

Actually lime (

C a O

) is added which in

H_{2} O

forms

C a (O H)_{2}

and precipitates

C a C O_{3}

C a O + H_{2} O \to C a (O H)_{2}

10

L of hard water requires

0.56

g of

C a O

, temporary hardness of

C a C O_{3}

x \times 10^{2}

ppm. Find

x

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10 L of hard water required 0.56 g of lime (CaO) for removing hardness. Hence, the temporary hardness in ppm (part per million) of CaCO3 is :

The correct option is B 200Temporary hardness is due to HCO⊖3 of Ca2+ and Mg2+.Ca(HCO3)2+CaO56g→2CaCO3(2×100)g+H2O0.56 g CaO≡2 g CaCO3 in 10 L H2O=2 g CaCO3 in 104 mL H2O=200 g CacO3 in 106 mL H2O

$10$ L of hard water required $0.56$ g of lime (CaO) for removing hardness. Hence, the temporary hardness in ppm (part per million) of $C a C O_{3}$ is :