Respuesta :
We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.
M1V1 = M2V2
0.266 M x V1 = 0.075 M x 150 mL
V1 = 42.29 mL
Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.
[tex]\boxed{{\text{42}}{\text{.3 mL}}}[/tex] of a 0.266 M [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution is required to make 150 mL of a 0.075 M [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.
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 molarity equation is given by the following expression:
[tex]{{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}[/tex] …… (1)
Here,
[tex]{{\text{M}}_{\text{1}}}[/tex] is the molarity of the initial [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.
[tex]{{\text{V}}_{_{\text{1}}}}[/tex] is the volume of the initial [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.
[tex]{{\text{M}}_{\text{2}}}[/tex] is the molarity of the new [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.
[tex]{{\text{V}}_{_{\text{2}}}}[/tex] is the volume of the new [tex]{\text{LiN}}{{\text{O}}_{\text{3}}}[/tex] solution.
Rearrange equation (1) to calculate [tex]{{\text{V}}_{\text{1}}}[/tex].
[tex]{{\text{V}}_{\text{1}}}=\frac{{{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{M}}_{\text{1}}}}}[/tex] …… (2)
The value of [tex]{{\text{M}}_{\text{1}}}[/tex] is 0.266 M.
The value of [tex]{{\text{M}}_{\text{2}}}[/tex] is 0.075 M.
The value of [tex]{{\text{V}}_{_{\text{2}}}}[/tex] is 150 mL.
Substitute these values in equation (2).
[tex]\begin{aligned}{{\text{V}}_{\text{1}}}&=\frac{{\left({{\text{0}}{\text{.075 M}}} \right)\left( {{\text{150 mL}}} \right)}}{{{\text{0}}{\text{.266 M}}}}\\&=42.29{\text{ mL}}\\&\approx 42.{\text{3 mL}}\\\end{aligned}[/tex]
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Concentration terms
Keywords: molarity, LiNO3, 42.3 mL, molarity equation, volume, M1, M2, V1, V2, 150 mL, 0.075 M, 0.266 M, concentration, concentration terms.
