Question

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

Answer 1

To prepare a 2.0M solution, you should add approximately 306 grams of Al(NO2)3 to 1.3L of water.

Step-by-step explanation:

To prepare a 2.0M solution, we can use the formula: Molarity (M) = moles of solute / volume of solution in liters. Rearranging the formula, moles of solute = Molarity x volume of solution. Given that the Molarity is 2.0M and the volume is 1.3L, we can calculate the moles of solute. Starting with the balanced equation Al(NO2)3 + 3H2O -> Al(OH)3 + 3HNO2, we can see that the ratio of moles of Al(NO2)3 to moles of H2O is 1:3. Therefore, 2 moles of Al(NO2)3 will react with 6 moles of H2O. Now we can calculate the moles of Al(NO2)3: 2.0M x 1.3L = 2.6 moles. So, you should add 2.6 moles or approximately 306 grams of Al(NO2)3 to 1.3L of water to prepare a 2.0M solution.

Answer 2

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

Step-by-step explanation:

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

`Molarity=(n)/(V_s)`

where,

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