Question

Calculate the number of moles of calcium ions present in a 50.00 mL water sample that has a hardness of 75.0 ppm (hardness due to CaCO3)

Answer 1

The number of moles of calcium ions present in a 50.00 mL water sample with hardness of 75.0 ppmis $\boxed{0.00003747{\text{ mol}}}$.

Further Explanation:

The hardness of water is calculated with the use of following formula:

${\text{Hardness}} = \dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}$                                                                      …… (1)

Since hardness in given problem occurs due to presence of ${\text{CaC}}{{\text{O}}_{\text{3}}}$, the above formula involves mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ and volume of ${\text{CaC}}{{\text{O}}_{\text{3}}}$.

Rearrange equation (1) to calculate mass.

${\text{Mass}} = \left( {{\text{Hardness}}} \right)\left( {{\text{Volume}}} \right)$                                                       …… (2)

In case of water, 1 ppm is approximately equivalent to 1 mg/L. Therefore hardness becomes 75.0 mg/L.

Volume is given in mL. It is to be converted into L. The conversion factor for this is,

$1{\text{ mL}} = {10^{ - 3}}{\text{ L}}$  

Therefore volume of water sample can be calculated as follows:

$\begin{aligned}{\text{Volume of water sample}} &= \left( {50.00{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 0.05{\text{ L}} \\\end{aligned}$  

Substitute 75.0 mg/L for hardness and 0.05 L for volume in equation (2) to calculate mass of ${\text{CaC}}{{\text{O}}_3}$.

$\begin{aligned}{\text{Mass of CaC}}{{\text{O}}_{\text{3}}} &= \left( {75.0{\text{ mg/L}}} \right)\left( {0.05{\text{ L}}} \right) \\&= 3.75{\text{ mg}} \\\end{aligned}$  

Mass of ${\text{CaC}}{{\text{O}}_3}$ is to be converted from mg to g. The conversion factor for this is,

$1{\text{ mg}} = {10^{ - 3}}{\text{ g}}$  

Therefore mass of ${\text{CaC}}{{\text{O}}_3}$ can be calculated as follows:

 $\begin{aligned}{\text{Mass of CaC}}{{\text{O}}_{\text{3}}} &= \left( {3.75{\text{ mg}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ g}}}}{{1{\text{ mg}}}}} \right) \\&= 0.00375{\text{ g}} \\\end{aligned}$

The formula to calculate moles of ${\text{CaC}}{{\text{O}}_3}$ is as follows:

${\text{Moles of CaC}}{{\text{O}}_{\text{3}}} = \dfrac{{{\text{Mass of CaC}}{{\text{O}}_{\text{3}}}}}{{{\text{Molar mass of CaC}}{{\text{O}}_{\text{3}}}}}$                                                    …… (3)

Substitute 0.00375 g for mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ and 100.08 g/mol for molar mass of    ${\text{CaC}}{{\text{O}}_{\text{3}}}$ in equation (3).

$\begin{aligned}{\text{Moles of CaC}}{{\text{O}}_{\text{3}}} &= \frac{{{\text{0}}{\text{.00375 g}}}}{{{\text{100}}{\text{.08 g/mol}}}} \\&= 0.00003747{\text{ mol}} \\\end{aligned}$  

Therefore number of calcium ions present in the given water sample is 0.00003747 mol.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603
2. Calculate the moles of ions in the solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: CaCO3, mass, volume, hardness, molar mass, moles, mass of CaCO3, 3.75 mg, 0.00375 g, 75.0 ppm, 50.00 mL, 0.00003747 mol

Answer 2

Hardness = 75ppm = 75 mg / L

volume = 50 mL =  0.05 L

So, applying hardness formula:

Hardness = mass / volume

so, mass =  hardness x volume = 75 x 0.05 =  3.75 mg = 0.00375 g

So, moles of CaCO3 = moles of Ca2+ ions = mass / molar mass of CaCO3 =  0.00375 / 100.06  =  0.00003747751 moles