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CHXI09:HYDROGEN

315529 $1000 g$ aqueous solution of ${CaCO}_{3}$ contains 10 $g$ of calcium carbonate. Hardness of solution is

1

10

p p m

2

100

p p m

3

1000

p p m

4

10000

p p m

Explanation:

Amount of ${CaCO}_{3}$ in ppm is $= \frac{10 \times 10^{6}}{1000} = 10^{4} ppm$ .

CHXI09:HYDROGEN

315530 $100 mL$ of tap water containing $Ca {({HCO}_{3})}_{2}$ was titrated with $N / 50$ $H C l$ with methyl orange as indicator. If $30 mL$ of $HCl$ were required, the temporary hardness as parts of ${CaCO}_{3}$ per $10^{6}$ parts of water is

1

150

p p m

2

300

p p m

3

450

p p m

4

600

p p m

Explanation:

$30$ $m L$ $N / 50$ $H C l$ $\equiv 30 m L$ $\equiv 30 m L$ $C a {(H C O_{3})}_{2}$
$\equiv 30 mL N/50 CaC O_{3} \equiv 100 mL$ tap water
Mass of ${CaCO}_{3}$ in $100 mL$ tap water
$= \frac{E \times N \times V}{1000}$
$= \frac{50 \times 30}{50 \times 1000} = 0.03 g \Rightarrow$ Hardness $300$ $p p m$

CHXI09:HYDROGEN

315531 Hardness of water is estimated by

1 EDTA method

2 Titrimetric method

3 conductivity method

4 distillation method

CHXI09:HYDROGEN

315532 The degree of hardness of water is usually expressed in terms of

1 ppm by weight of

{MgSO}_{4}

2

g / L

of

{CaCO}_{3}

and

{MgCO}_{3}

present

3 ppm by weight of

{CaCO}_{3}

irrespective of whether it is actually present

4 ppm of

{CaCO}_{3}

actually present in water

Explanation:

Degree of hardness of water is measured in terms of ppm by weight of $C a C o_{3}$ irrespective of whether it is actually present or not.

CHXI09:HYDROGEN

315529 $1000 g$ aqueous solution of ${CaCO}_{3}$ contains 10 $g$ of calcium carbonate. Hardness of solution is

1

10

p p m

2

100

p p m

3

1000

p p m

4

10000

p p m

Explanation:

Amount of ${CaCO}_{3}$ in ppm is $= \frac{10 \times 10^{6}}{1000} = 10^{4} ppm$ .

CHXI09:HYDROGEN

315530 $100 mL$ of tap water containing $Ca {({HCO}_{3})}_{2}$ was titrated with $N / 50$ $H C l$ with methyl orange as indicator. If $30 mL$ of $HCl$ were required, the temporary hardness as parts of ${CaCO}_{3}$ per $10^{6}$ parts of water is

1

150

p p m

2

300

p p m

3

450

p p m

4

600

p p m

Explanation:

$30$ $m L$ $N / 50$ $H C l$ $\equiv 30 m L$ $\equiv 30 m L$ $C a {(H C O_{3})}_{2}$
$\equiv 30 mL N/50 CaC O_{3} \equiv 100 mL$ tap water
Mass of ${CaCO}_{3}$ in $100 mL$ tap water
$= \frac{E \times N \times V}{1000}$
$= \frac{50 \times 30}{50 \times 1000} = 0.03 g \Rightarrow$ Hardness $300$ $p p m$

CHXI09:HYDROGEN

315531 Hardness of water is estimated by

1 EDTA method

2 Titrimetric method

3 conductivity method

4 distillation method

CHXI09:HYDROGEN

315532 The degree of hardness of water is usually expressed in terms of

1 ppm by weight of

{MgSO}_{4}

2

g / L

of

{CaCO}_{3}

and

{MgCO}_{3}

present

3 ppm by weight of

{CaCO}_{3}

irrespective of whether it is actually present

4 ppm of

{CaCO}_{3}

actually present in water

Explanation:

Degree of hardness of water is measured in terms of ppm by weight of $C a C o_{3}$ irrespective of whether it is actually present or not.

CHXI09:HYDROGEN

315529 $1000 g$ aqueous solution of ${CaCO}_{3}$ contains 10 $g$ of calcium carbonate. Hardness of solution is

1

10

p p m

2

100

p p m

3

1000

p p m

4

10000

p p m

Explanation:

Amount of ${CaCO}_{3}$ in ppm is $= \frac{10 \times 10^{6}}{1000} = 10^{4} ppm$ .

CHXI09:HYDROGEN

315530 $100 mL$ of tap water containing $Ca {({HCO}_{3})}_{2}$ was titrated with $N / 50$ $H C l$ with methyl orange as indicator. If $30 mL$ of $HCl$ were required, the temporary hardness as parts of ${CaCO}_{3}$ per $10^{6}$ parts of water is

1

150

p p m

2

300

p p m

3

450

p p m

4

600

p p m

Explanation:

$30$ $m L$ $N / 50$ $H C l$ $\equiv 30 m L$ $\equiv 30 m L$ $C a {(H C O_{3})}_{2}$
$\equiv 30 mL N/50 CaC O_{3} \equiv 100 mL$ tap water
Mass of ${CaCO}_{3}$ in $100 mL$ tap water
$= \frac{E \times N \times V}{1000}$
$= \frac{50 \times 30}{50 \times 1000} = 0.03 g \Rightarrow$ Hardness $300$ $p p m$

CHXI09:HYDROGEN

315531 Hardness of water is estimated by

1 EDTA method

2 Titrimetric method

3 conductivity method

4 distillation method

CHXI09:HYDROGEN

315532 The degree of hardness of water is usually expressed in terms of

1 ppm by weight of

{MgSO}_{4}

2

g / L

of

{CaCO}_{3}

and

{MgCO}_{3}

present

3 ppm by weight of

{CaCO}_{3}

irrespective of whether it is actually present

4 ppm of

{CaCO}_{3}

actually present in water

Explanation:

Degree of hardness of water is measured in terms of ppm by weight of $C a C o_{3}$ irrespective of whether it is actually present or not.

CHXI09:HYDROGEN

315529 $1000 g$ aqueous solution of ${CaCO}_{3}$ contains 10 $g$ of calcium carbonate. Hardness of solution is

1

10

p p m

2

100

p p m

3

1000

p p m

4

10000

p p m

Explanation:

Amount of ${CaCO}_{3}$ in ppm is $= \frac{10 \times 10^{6}}{1000} = 10^{4} ppm$ .

CHXI09:HYDROGEN

315530 $100 mL$ of tap water containing $Ca {({HCO}_{3})}_{2}$ was titrated with $N / 50$ $H C l$ with methyl orange as indicator. If $30 mL$ of $HCl$ were required, the temporary hardness as parts of ${CaCO}_{3}$ per $10^{6}$ parts of water is

1

150

p p m

2

300

p p m

3

450

p p m

4

600

p p m

Explanation:

$30$ $m L$ $N / 50$ $H C l$ $\equiv 30 m L$ $\equiv 30 m L$ $C a {(H C O_{3})}_{2}$
$\equiv 30 mL N/50 CaC O_{3} \equiv 100 mL$ tap water
Mass of ${CaCO}_{3}$ in $100 mL$ tap water
$= \frac{E \times N \times V}{1000}$
$= \frac{50 \times 30}{50 \times 1000} = 0.03 g \Rightarrow$ Hardness $300$ $p p m$

CHXI09:HYDROGEN

315531 Hardness of water is estimated by

1 EDTA method

2 Titrimetric method

3 conductivity method

4 distillation method

CHXI09:HYDROGEN

315532 The degree of hardness of water is usually expressed in terms of

1 ppm by weight of

{MgSO}_{4}

2

g / L

of

{CaCO}_{3}

and

{MgCO}_{3}

present

3 ppm by weight of

{CaCO}_{3}

irrespective of whether it is actually present

4 ppm of

{CaCO}_{3}

actually present in water

Explanation:

Degree of hardness of water is measured in terms of ppm by weight of $C a C o_{3}$ irrespective of whether it is actually present or not.

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