Respuesta :
Answer:
Molarity=0.0058M
Explanation:
Given the weight of AgN[tex]O_{3}[/tex] is(w) 1.24g
We know that the molecular weight of AgN[tex]O_{3}[/tex] is(W) 169.87g/mol.
Given the volume of the solution is(V) 125ml.
Here solute is silver nitrate and the solvent is water.
We know that
[tex]Molarity=\frac{number of moles of solute}{total volume of the solution int litres}[/tex]
We know that number of moles=weight/molecular weight
[tex]n= \frac{1.24}{169.87}[/tex]= [tex]7.3*10^{-3}[/tex]
volume of the solution(V)=125/1000=0.125L
Now
Molarity=n/V
M= [tex]\frac{7.3*10^{-3} }{0.125}[/tex] =0.00584M
0.058 mol/L
Explanation:
[tex]\textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}[/tex]
We are given a [tex]125mL[/tex] solution which contains [tex]1.24g[/tex] of [tex]AgNO_{3}[/tex].
To calculate number of moles of [tex]AgNO_{3}[/tex] present, we need to find it's molar mass from charts. Molar Mass of [tex]AgNO_{3}[/tex] is known to be [tex]169.87\frac{g}{Mol}[/tex].
Number of moles of [tex]AgNO_{3}[/tex] present = [tex]\frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{1.24g}{169.87\frac{g}{Mol}}\textrm{ = }0.0073mol[/tex]
Molarity = [tex]\frac{0.0073mol}{0.125L}=0.0584\frac{mol}{L}[/tex]
∴ Molarity of given [tex]AgNO_{3}[/tex] solution = [tex]0.058\frac{mol}{L}[/tex]
