Question

How many grams of Al(no2)3 should be added to 1.3L of water to prepare a 2.0M solution

Answer 1

Answer: 429 grams of $Al(NO_2)_3$  should be added to 1.3 L of water to prepare 2.0 M solution

Explanation:

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

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

where,

n = moles of solute $Al(NO_2)_3$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{165g/mol}$ 

$V_s$ = volume of solution in Liters  

Now put all the given values in the formula of molarity, we get

$2.0M=\frac{xg}{165g/mol\times 1.3L}$

$x=429g$

Therefore, 429 grams of $Al(NO_2)_3$  should be added to 1.3 L of water to prepare 2.0 M solution