Question

How many mL of a 0.63 M solution would contain 12g of Al(NO3)3

Answer 1

Answer : 89 mL of solution would contain the given amount of Al(NO₃)₃.

Explanation :

Step 1 : Find moles of Al(NO₃)₃.

The molar mass of Al(NO₃)₃ is 213 g/mol

The formula to calculate mole is given below.

`Mole = (Mass (grams))/(MolarMass)`

We have 12 g of Al(NO₃)₃. Let us plug in this value to find mol.

`mole = (12g)/(213 g/mol)`

Mole = 0.056 mol.

We have 0.056 mols of Al(NO₃)₃

Step 2 : Use molarity formula to find the volume.

The molarity of a solution is defined as moles of solute per liter of solution.

This can be represented in terms of formula as follows.

`Molarity (M)= (mol)/(L)`

We have 0.63 M solution.

`0.63 M = (0.056mol)/(L)`

On rearranging we get,

`L = (0.056)/(0.63) = 0.089`

We have 0.089 L of solution. Let us convert this to mL.

`0.089 L * (1000mL)/(1L) = 89 mL`

89 mL of solution would contain the given amount of Al(NO₃)₃.