Question

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

Answer 1

0.058 mol/L

Step-by-step 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(g)/(Mol)`.

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

Molarity =
`(0.0073mol)/(0.125L)=0.0584(mol)/(L)`

∴ Molarity of given
`AgNO_(3)` solution =
`0.058(mol)/(L)`

Answer 2

Molarity=0.0058M

Step-by-step 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=(number of moles of solute)/(total volume of the solution int litres)`

We know that number of moles=weight/molecular weight

`n= (1.24)/(169.87)`=
`7.3*10^(-3)`

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

Now

Molarity=n/V

M=
`(7.3*10^(-3) )/(0.125)` =0.00584M