Answer
Verified452.1k+ views
Hint:To calculate the volume of $N{a_2}C{O_3}$, the formula for calculating normality and milliequivalent is used. The millie equivalent is calculated in milligrams. The normality is defined as the gram equivalent mass of compound present in one litre solution.
Complete answer:Given
Volume of pond water is 1 L.
Mass of $C{a^{2 + }}$ ion is 20 mg.
Mass of $M{g^{2 + }}$ ion is 12 mg.
Normality of $N{a_2}C{O_3}$ is 2N.
$C{a^{2 + }}$ ion forms calcium carbonate and $M{g^{2 + }}$ ion forms magnesium carbonate.
Milliequivalent of $C{a^{2 + }}$ ion and milliequivalent of $M{g^{2 + }}$ ion is equal to the milli equivalent of $N{a_2}C{O_3}$.
The equation is given as shown below
Milli eq. of $C{a^{2 + }}$+ Milli eq. of $M{g^{2 + }}$= Milli.eq. of $N{a_2}C{O_3}$……(i)
Milliequivalent is defined as the mass in milligrams of the solute component equal to 1/1000 of its gram equivalent mass by considering the valency of the ions.
The formula for calculating the milli equivalent is shown below.
$mEq = \dfrac{{m \times V}}{M}$
Where,
mEq is the millie equivalent.
m is the mass in mg.
V is the valency
M is the molecular weight.
Normality is defined as the concentration of solution measured in terms of gram equivalent of solute in one litre solution or milliequivalent of solute is one litre solution.
The formula for normality is shown below.
$N = \dfrac{{meq}}{V}$
Where,
N is the normality.
meq is milliequivalent
V is the volume.
To calculate the volume $N{a_2}C{O_3}$ solution, substitute the values, in equation (i).
$\Rightarrow \dfrac{{20}}{{40}} \times 2 + \dfrac{{12}}{{24}} \times 2 = 2 \times V$
$\Rightarrow 0.5 \times 2 + 0.5 \times 2 = 2 \times V$
$\Rightarrow 1 + 1 = 2 \times V$
$\Rightarrow V = 1mL$
1mL of $N{a_2}C{O_3}$ solution is used to soften 1 L of pond water.
To soften 5000 L of pond water, the volume needed is
$\Rightarrow 1 \times {10^{ - 3}} \times 5000$
$\Rightarrow 5L$
Therefore, the correct option is C.
Note:
Make sure to convert volume in millilitre to litre as the volume is calculated in terms of litre. Calcium carbonate and magnesium carbonate present in the pond water is insoluble in water, thus it is removed from the water to form soft water.
Complete answer:Given
Volume of pond water is 1 L.
Mass of $C{a^{2 + }}$ ion is 20 mg.
Mass of $M{g^{2 + }}$ ion is 12 mg.
Normality of $N{a_2}C{O_3}$ is 2N.
$C{a^{2 + }}$ ion forms calcium carbonate and $M{g^{2 + }}$ ion forms magnesium carbonate.
Milliequivalent of $C{a^{2 + }}$ ion and milliequivalent of $M{g^{2 + }}$ ion is equal to the milli equivalent of $N{a_2}C{O_3}$.
The equation is given as shown below
Milli eq. of $C{a^{2 + }}$+ Milli eq. of $M{g^{2 + }}$= Milli.eq. of $N{a_2}C{O_3}$……(i)
Milliequivalent is defined as the mass in milligrams of the solute component equal to 1/1000 of its gram equivalent mass by considering the valency of the ions.
The formula for calculating the milli equivalent is shown below.
$mEq = \dfrac{{m \times V}}{M}$
Where,
mEq is the millie equivalent.
m is the mass in mg.
V is the valency
M is the molecular weight.
Normality is defined as the concentration of solution measured in terms of gram equivalent of solute in one litre solution or milliequivalent of solute is one litre solution.
The formula for normality is shown below.
$N = \dfrac{{meq}}{V}$
Where,
N is the normality.
meq is milliequivalent
V is the volume.
To calculate the volume $N{a_2}C{O_3}$ solution, substitute the values, in equation (i).
$\Rightarrow \dfrac{{20}}{{40}} \times 2 + \dfrac{{12}}{{24}} \times 2 = 2 \times V$
$\Rightarrow 0.5 \times 2 + 0.5 \times 2 = 2 \times V$
$\Rightarrow 1 + 1 = 2 \times V$
$\Rightarrow V = 1mL$
1mL of $N{a_2}C{O_3}$ solution is used to soften 1 L of pond water.
To soften 5000 L of pond water, the volume needed is
$\Rightarrow 1 \times {10^{ - 3}} \times 5000$
$\Rightarrow 5L$
Therefore, the correct option is C.
Note:
Make sure to convert volume in millilitre to litre as the volume is calculated in terms of litre. Calcium carbonate and magnesium carbonate present in the pond water is insoluble in water, thus it is removed from the water to form soft water.