Question

Calculate the molarity when 1.24 g AgNO3 is dissolved to make 125 mL solution. Round to two significant digits.

Answer 1

Answer:

Molarity=0.0058M

Explanation:

Given the weight of AgN$O_{3}$ is(w) 1.24g

We know that the molecular weight of AgN$O_{3}$ 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

$Molarity=\frac{number of moles of solute}{total volume of the solution int litres}$

We know that number of moles=weight/molecular weight

$n= \frac{1.24}{169.87}$=   $7.3*10^{-3}$

volume of the solution(V)=125/1000=0.125L

Now

Molarity=n/V

M=  $\frac{7.3*10^{-3} }{0.125}$   =0.00584M

Answer 2

0.058 mol/L

Explanation:

       $\textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}$

       We are given a $125mL$ solution which contains $1.24g$ of $AgNO_{3}$.

       To calculate number of moles of $AgNO_{3}$ present, we need to find it's molar mass from charts. Molar Mass of $AgNO_{3}$ is known to be $169.87\frac{g}{Mol}$.

       Number of moles of $AgNO_{3}$ present = $\frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{1.24g}{169.87\frac{g}{Mol}}\textrm{ = }0.0073mol$

       Molarity = $\frac{0.0073mol}{0.125L}=0.0584\frac{mol}{L}$

∴ Molarity of given $AgNO_{3}$ solution = $0.058\frac{mol}{L}$