Respuesta :
Answer: [tex]5.70\times 10^{-6}M[/tex]
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
Given : 0.360 g of [tex]KNO_3[/tex] is dissolved in 500 ml of solution.
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.360g}{101g/mol}=3.56\times 10^{-3}mole[/tex]
[tex]V_s[/tex] = volume of solution = 500 ml
[tex]Molarity=\frac{3.56\times 10^{-3}\times 1000}{500}=7.12\times 10^{-3}M[/tex]
According to the neutralization law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = [tex]7.12\times 10^{-3}M[/tex]
[tex]V_1[/tex] = volume of stock solution = 10.0 ml
[tex]M_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 500.0 ml
[tex]7.12\times 10^{-3}M\times 10.0=M_2\times 500.0[/tex]
[tex]M_2=1.42\times 10^{-4}M[/tex]
b) On further dilution
[tex]M_1[/tex] = molarity of stock solution = [tex]1.42\times 10^{-4}M[/tex]
[tex]V_1[/tex] = volume of stock solution = 10.0 ml
[tex]M_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 250.0 ml
[tex]1.42\times 10^{-4}M\times 10.0=M_2\times 250.0[/tex]
[tex]M_2=5.70\times 10^{-6}M[/tex]
Thus the final concentration of the [tex]KNO_3[/tex] solution is [tex]5.70\times 10^{-6}M[/tex]
