Question

a) What mass of aluminium nitrate (Al(NO3)3) would be required to prepare 2.000 L of a 0.0170 M aqueous solution of this salt

Answer 1

Answer: 7.24 g of aluminium nitrate is required to prepare 2.000 L of a 0.0170 M aqueous solution of this salt.

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Given :

Molarity of solution = 0.0170 M

Volume of solution = 2.000 L

$Molarity=\frac{n}{V_s}$

where,

n= moles of solute

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{213g/mol}$

$V_s$ = volume of solution in Liter = 2.000 L

$0.0170M=\frac{x}{213g/mol\times 2.000L}$

$x=7.24g$

7.24 g of aluminium nitrate is required to prepare 2.000 L of a 0.0170 M aqueous solution of this salt.