Answer:
[tex]\boxed{\text{0.118 mol/L}}[/tex]
Explanation:
[tex]\text{Molar concentration } = \dfrac{\text{moles}}{\text{litres}}\\\\\text{c} = \dfrac{n}{V}[/tex]
1. Convert grams to moles.
[tex]\text{Moles AgNO}_{3} = \text{10.0 g AgNO}_{3} \times \dfrac{\text{1 mol AgNO}_{3} }{\text{169.87g AgNO}_{3} } = \text{0.058 87 mol AgNO}_{3}[/tex]
2. Convert moles to litres
[tex]\text{V = 500. mL} \times \dfrac{\text{1 L}}{\text{1000 mL}} = \text{0.5000 L}[/tex]
3. Calculate the molar concentration
[tex]c = \dfrac{\text{0.058 87 mol}}{\text{0.5000 L}} = \text{0.118 mol/L}[/tex]
The molar concentration of AgNO₃ is [tex]\boxed{\textbf{0.118 mol/L}}[/tex].
The molarity of a solution in which 10.0g of AgNO₃ is dissolved in 500. mL of solution is 20M.
Molarity of any solution is define as the number of moles of solute present in per liter of solution.
M = n/V, where
n = moles of solute AgNO₃ = 10g
V = volume of solution = 500mL = 0.5L
On putting values on the above equation, we get
M = 10/0.5 = 20M
Hence required molarity of solution is 20 M.
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