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

Sagot :

Answer: 429 grams of [tex]Al(NO_2)_3[/tex]  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.

[tex]Molarity=\frac{n}{V_s}[/tex]

where,

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

[tex]V_s[/tex] = volume of solution in Liters  

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

[tex]2.0M=\frac{xg}{165g/mol\times 1.3L}[/tex]

[tex]x=429g[/tex]

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

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