Respuesta :
Answer:
For 1: The molarity of [tex]K^+\text{ ions}[/tex] in this solution is 0.0592 M
For 2: The new molarity of [tex]K^+\text{ ions}[/tex] in this solution is [tex]2.07\times 10^{-3}M[/tex]
Explanation:
- For 1:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of KOH = 1.66 g
Molar mass of KOH = 56.1 g/mol
Volume of solution = 500.0 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{1.66\times 1000}{56.1g/mol\times 500.0}\\\\\text{Molarity of solution}=0.0592M[/tex]
1 mole of KOH produces 1 mole of potassium ions and 1 mole of hydroxide ions
So, molarity of [tex]K^+\text{ ions}=0.0592M[/tex]
Hence, the molarity of [tex]K^+\text{ ions}[/tex] in this solution is 0.0592 M
- For 2:
To calculate the molarity of the diluted solution, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated KOH solution having [tex]K^+\text{ ions}[/tex]
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted KOH solution having [tex]K^+\text{ ions}[/tex]
We are given:
[tex]M_1=0.0592M\\V_1=35.00mL\\M_2=?M\\V_2=1000mL[/tex]
Putting values in above equation, we get:
[tex]0.0592\times 35.00=M_2\times 1000\\\\M_2=\frac{0.0592\times 35.0}{1000}=2.07\times 10^{-3}M[/tex]
Hence, the new molarity of [tex]K^+\text{ ions}[/tex] in this solution is [tex]2.07\times 10^{-3}M[/tex]
