
How many grams of ${\text{MgC}}{{\text{l}}_{\text{2}}}$ would be required to produce a \[4.5{\text{ M}}\] solution with a volume of 2L?
Answer
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Hint: We are already provided with the compound. We know that a 1-M solution contains 1 mole of solute for every 1 L of solution. Then calculate the moles of the reactants to further calculate the weight of the compound.
Formula used: \[{\text{moles = }}\dfrac{{{\text{ given mass }}}}{{{\text{molar mass}}}}\]
${\text{molarity}} = \dfrac{{{\text{moles}}}}{{{\text{volume of solution}}}}$
Complete step-by-step answer:
We already know that a \[4.5{\text{ M}}\] magnesium chloride solution will contain \[4.5\] moles of magnesium chloride, the solute, for every 1 L of solution.
So all we have to do now is figure out how many moles of magnesium chloride would be equivalent in 2 L of solution to \[4.5\] moles in 1.0L of solution.
If the sample was measured in grams, we would first need to convert it to moles and then determine the amount of magnesium chloride.
${\text{moles }}\left( {{\text{MgC}}{{\text{l}}_{\text{2}}}} \right){\text{ = }}\dfrac{{{\text{4}}{\text{.5 moles}}}}{{{\text{1}}{\text{.0 L}}}}{{ \times 2}}{\text{.0 L }}\left( {{\text{solution}}} \right)$
${\text{ = 9}}{\text{.0 moles of MgC}}{{\text{l}}_{\text{2}}}$
So, you know that you can get a \[4.5{\text{ M}}\] magnesium chloride solution by dissolving 9 moles of magnesium chloride in enough water to have a total volume of 2 L of solution.
To convert this to grams, use the compound's molar mass
${\text{wt }}\left( {{\text{MgC}}{{\text{l}}_{\text{2}}}} \right){\text{ = 9}}{{.0 mol \times 95}}{\text{.211 g}}$
${\text{ = 860g of MgC}}{{\text{l}}_{\text{2}}}$
So, the weight of ${\text{MgC}}{{\text{l}}_{\text{2}}}$ is 860g.
Additional Information: One mole of any substance is equal to the value of \[6.023 \times {10^{23}}\] (Avagadro number). It can be used to measure the products obtained from the chemical reaction. The unit is denoted by mol.
Note:We already know that the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of the compound by the molar mass of the compound expressed in grams.
All the values should be put carefully and the equations should be solved neatly.
Formula used: \[{\text{moles = }}\dfrac{{{\text{ given mass }}}}{{{\text{molar mass}}}}\]
${\text{molarity}} = \dfrac{{{\text{moles}}}}{{{\text{volume of solution}}}}$
Complete step-by-step answer:
We already know that a \[4.5{\text{ M}}\] magnesium chloride solution will contain \[4.5\] moles of magnesium chloride, the solute, for every 1 L of solution.
So all we have to do now is figure out how many moles of magnesium chloride would be equivalent in 2 L of solution to \[4.5\] moles in 1.0L of solution.
If the sample was measured in grams, we would first need to convert it to moles and then determine the amount of magnesium chloride.
${\text{moles }}\left( {{\text{MgC}}{{\text{l}}_{\text{2}}}} \right){\text{ = }}\dfrac{{{\text{4}}{\text{.5 moles}}}}{{{\text{1}}{\text{.0 L}}}}{{ \times 2}}{\text{.0 L }}\left( {{\text{solution}}} \right)$
${\text{ = 9}}{\text{.0 moles of MgC}}{{\text{l}}_{\text{2}}}$
So, you know that you can get a \[4.5{\text{ M}}\] magnesium chloride solution by dissolving 9 moles of magnesium chloride in enough water to have a total volume of 2 L of solution.
To convert this to grams, use the compound's molar mass
${\text{wt }}\left( {{\text{MgC}}{{\text{l}}_{\text{2}}}} \right){\text{ = 9}}{{.0 mol \times 95}}{\text{.211 g}}$
${\text{ = 860g of MgC}}{{\text{l}}_{\text{2}}}$
So, the weight of ${\text{MgC}}{{\text{l}}_{\text{2}}}$ is 860g.
Additional Information: One mole of any substance is equal to the value of \[6.023 \times {10^{23}}\] (Avagadro number). It can be used to measure the products obtained from the chemical reaction. The unit is denoted by mol.
Note:We already know that the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of the compound by the molar mass of the compound expressed in grams.
All the values should be put carefully and the equations should be solved neatly.
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