Respuesta :
Answer:
Concentration of calcium chloride solution is 0.5623 molar.
Explanation:
Density of the solution = 1.05 g/mL
Volume of the solution = V
Mass of the solution = Mass of solute +Mass of solvent
= 23.7 g + 375 g = 398.7 g
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]V= \frac{398.7 g}{1.05 g/mL}=379.71 mL=0.37971 L[/tex]
Moles of solute that id calcium chloride =[tex]\frac{23.7 g}{111 g/mol}=0.2135 mol[/tex]
Concentration=[tex]\frac{\text{moles of solute}}{\text{Volume of the solution in L}}[/tex]
Concentration of calcium chloride solution is;
[tex][CaCl_2]=\frac{0.2135 mol}{0.37971 L}=0.5623 molar[/tex]
The molarity of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] solution is [tex]\boxed{{\text{0}}{\text{.5623}}\;{\text{M}}}[/tex].
Further Explanation:
The concentration is the proportion of substance in the mixture. The most commonly used concentration terms are as follows:
1. Molarity (M)
2. Molality (m)
3. Mole fraction (X)
4. Parts per million (ppm)
5. Mass percent ((w/w) %)
6. Volume percent ((v/v) %)
Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M, and its unit is mol/L.
The formula to calculate the molarity of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] solution is as follows:
[tex]{\text{Molarity of CaC}}{{\text{l}}_{\text{2}}}{\text{ solution}} =\dfrac{{{\text{amount}}\;\left( {{\text{mol}}} \right)\;{\text{of}}\;{\text{CaC}}{{\text{l}}_{\text{2}}}}}{{\;{\text{volume }}\left( {\text{L}} \right)\;{\text{of}}\;{\text{CaC}}{{\text{l}}_{\text{2}}}{\text{ solution}}}}[/tex] ......(1)
To calculate the volume of total [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] solution, firstly we calculate the total mass of the solution.
The formula to calculate the total mass of solution is:
[tex]{\text{Total mass of}}\;{\text{solution }} = {\text{Mass of\;CaC}}{{\text{l}}_{\text{2}}} + {\text{Mass of\;}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] ......(2)
Substitute 23.7 g for mass of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] and 375 g for mass of [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] in equation (2).
[tex]\begin{aligned}{\text{Total mass of}}\;{\text{CaC}}{{\text{l}}_{\text{2}}}\;{\text{solution }} &= 23.7\;{\text{g}} + 375\;{\text{g}}\\&= 398.7\;{\text{g}}\\\end{aligned}[/tex]
Density of a substance is given by mass per unit volume of that substance. It is given by formula:
[tex]{\text{Density}} = \dfrac{{{\text{mass}}}}{{{\text{volume}}}}[/tex] ......(3)
Rearrange equation (3) to calculate volume of solution.
[tex]{\text{Volume}} = \dfrac{{{\text{Mass}}}}{{{\text{Density}}}}[/tex]
Substitute [tex]398.7\;{\text{g}}[/tex] for mass and [tex]1.05\;{\text{g/mL}}[/tex] for density in equation(3).
[tex]\begin{aligned}{\text{Volume}}&=\frac{{398.7\;{\text{g}}}}{{{\text{1}}{\text{.05}}\;{\text{g/mL}}}}\\&= {\text{379}}{\text{.71 mL}}\\\end{aligned}[/tex]
The volume is to be converted into L. The conversion factor for this is,
[tex]{\text{1 mL}} = {\text{1}}{{\text{0}}^{ - {\text{3}}}}{\text{ L}}[/tex]
So the volume of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] solution can be calculated as follows:
[tex]\begin{aligned}{\text{Volume of CaC}}{{\text{l}}_{\text{2}}}{\text{ solution}} &= \left( {{\text{379}}{\text{.71 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.37971\;{\text{L}}\\\end{aligned}[/tex]
The number of moles of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] can be determined as follows:
[tex]{\text{Moles of CaC}}{{\text{l}}_{\text{2}}} &= \dfrac{{{\text{Given mass of CaC}}{{\text{l}}_{\text{2}}}}}{{{\text{Molar mass of CaC}}{{\text{l}}_{\text{2}}}}}[/tex]
......(4)
Substitute [tex]23.7{\text{ g}}[/tex] for given mass of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] and [tex]111{\text{ g/mol}}[/tex] for molar mass of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] in equation (4).
[tex]\begin{aligned}{\text{Moles of CaC}}{{\text{l}}_{\text{2}}} &= \frac{{23.7\;{\text{g}}}}{{111\;{\text{g}}/{\text{mol}}}}\\&= 0.2135\;{\text{mol}}\\\end{aligned}[/tex]
The moles of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] is [tex]0.2135\;{\text{mol}}[/tex].
The volume of [tex]{\text{CaC}}{{\text{l}}_{\text{2}}}[/tex] solution is [tex]0.37971\;{\text{L}}[/tex].
Substitute these values in equation (1).
[tex]\begin{aligned}{\text{Molarity of CaC}}{{\text{l}}_{\text{2}}}{\text{ solution}} &= \frac{{0.2135{\text{ mol}}}}{{0.37971\;{\text{L}}}}\\&= 0.5623\;{\text{mol/L}}\\\end{aligned}[/tex]
So the molarity of [tex]{\mathbf{CaC}}{{\mathbf{l}}_{\mathbf{2}}}[/tex] solution is [tex]{\mathbf{0}}{\mathbf{.5623}}\;{\mathbf{mol/L}}[/tex].
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Concentration terms
Keywords: molarity, concentration, 0.5623M, 0.2135mol, 23.7g, 375g, 1.05g, density, mass, solution, CaCl2 and H2O.
