## Calculating a mass to be weighed

### For a mass concentration

Example: prepare a 100 mL solution of 1 mg/L of Na+ with NaCl salt

1 mg NaCl ≠ 1 mg Na+

MNaCl = 58.44 g/mol ; MNa = 22.99 g/mol ; MCl = 35.45 g/mol $\large \Rightarrow \% M{_{Na}} = \frac{M{_{Na}}}{M{_{NaCl}}} = 39.3\%$

It can be seen here that in 1 mg of NaCl, there is in fact only 0.39 mg of Na+. This ratio must therefore be taken into account when calculating the mass of salt to be weighed in order to obtain a concentration of 1mg/L of Na+.

#### Calculating the mass of salt to be weighed

$\LARGE m{_{salt}} = \frac{M{_{salt}}}{M{_{element}}} \times C{_{f}} \times V{_{f}}$
Where
msalt = mass of salt to be weighed (in g)
Msalt = molar mass of salt (in g/mol)
Melement = molar mass of the element to consider (in g/mol)
Cf = final mass concentration (in g/L)
Vf = final volume (in L)
##### Mass of NaCl to be weighed

$\large m{_{NaCl}} = \frac{58.44}{22.99} \times 10{^{-3}} \times 0.1 = 2.54 \times 10{^{-4}}g = \mathbf{{\color{DarkGreen} 0.254mg}}$

Thus in order to obtain a 100 mL solution of 1mg/L of Na+, we will need to weigh 0.254 mg of salt. In reality, weighing 0.254 mg of salt is practically impossible because the most precise scales have a precision of 0.1 mg. We will need to prepare a more concentrated solution and then dilute it in order to obtain the desired concentration level of Na+.

### For a molar concentration

Example: prepare a 100 mL solution of 1 mol/L Na+ with Na2SO4 salt

Pay attention to the ratio between the number of moles of salt and of ion.
Here 1 mole of Na2SO4 contains 2 moles of Na+
$\LARGE m{_{salt}} = \frac{n{_{salt}}}{n{_{element}}} \times C{_{f}} \times M{_{salt}} \times V{_{f}}$
Where
msalt = mass of salt to be weighed (in g)
Msalt = molar mass of salt (in g/mol)
nsalt/nelement = molar ratio between the element and the salt
Cf = final molar concentration (in mol/L)
Vf = final volume (in L)

Thus, the mass of salt to be weighed in order to prepare a 1 mol/L solution of Na+ is :

$\large m{_{NaCl}} = \frac{1}{2} \times 1 \times 142.04 \times 0.1 = \mathbf{{\color{DarkGreen} 7.1g}}$

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The text passage provides detailed examples of how to calculate the mass of salt needed to prepare solutions of specific concentrations, both in terms of mass concentration and molar concentration. It includes step-by-step calculations and provides the final answers for each example.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## Calculating a mass to be weighed

### For a mass concentration

Example: prepare a 100 mL solution of 1 mg/L of Na+ with NaCl salt

1 mg NaCl ≠ 1 mg Na+

MNaCl = 58.44 g/mol ; MNa = 22.99 g/mol ; MCl = 35.45 g/mol $\large \Rightarrow \% M{_{Na}} = \frac{M{_{Na}}}{M{_{NaCl}}} = 39.3\%$

It can be seen here that in 1 mg of NaCl, there is in fact only 0.39 mg of Na+. This ratio must therefore be taken into account when calculating the mass of salt to be weighed in order to obtain a concentration of 1mg/L of Na+.

#### Calculating the mass of salt to be weighed

$\LARGE m{_{salt}} = \frac{M{_{salt}}}{M{_{element}}} \times C{_{f}} \times V{_{f}}$
Where
msalt = mass of salt to be weighed (in g)
Msalt = molar mass of salt (in g/mol)
Melement = molar mass of the element to consider (in g/mol)
Cf = final mass concentration (in g/L)
Vf = final volume (in L)
##### Mass of NaCl to be weighed

$\large m{_{NaCl}} = \frac{58.44}{22.99} \times 10{^{-3}} \times 0.1 = 2.54 \times 10{^{-4}}g = \mathbf{{\color{DarkGreen} 0.254mg}}$

Thus in order to obtain a 100 mL solution of 1mg/L of Na+, we will need to weigh 0.254 mg of salt. In reality, weighing 0.254 mg of salt is practically impossible because the most precise scales have a precision of 0.1 mg. We will need to prepare a more concentrated solution and then dilute it in order to obtain the desired concentration level of Na+.

### For a molar concentration

Example: prepare a 100 mL solution of 1 mol/L Na+ with Na2SO4 salt

Pay attention to the ratio between the number of moles of salt and of ion.
Here 1 mole of Na2SO4 contains 2 moles of Na+
$\LARGE m{_{salt}} = \frac{n{_{salt}}}{n{_{element}}} \times C{_{f}} \times M{_{salt}} \times V{_{f}}$
Where
msalt = mass of salt to be weighed (in g)
Msalt = molar mass of salt (in g/mol)
nsalt/nelement = molar ratio between the element and the salt
Cf = final molar concentration (in mol/L)
Vf = final volume (in L)

Thus, the mass of salt to be weighed in order to prepare a 1 mol/L solution of Na+ is :

$\large m{_{NaCl}} = \frac{1}{2} \times 1 \times 142.04 \times 0.1 = \mathbf{{\color{DarkGreen} 7.1g}}$

Next page
Previous page
user
The text passage provides detailed examples of how to calculate the mass of salt needed to prepare solutions of specific concentrations, both in terms of mass concentration and molar concentration. It includes step-by-step calculations and provides the final answers for each example.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS