Question

How many grams would you need to make a 0.2 M solution of ALN at a volume of 0.20 L?

a) 7.2 g
b) 1.8 g
c) 36 g
d) 0.36 g

Answer 1

To make a 0.2 M solution of AlN with a volume of 0.20 L, one would need 1.8 g of AlN. This is calculated using the molar mass of AlN (40.99 g/mol) and the desired molarity and volume of the solution.

Step-by-step explanation:

The student is asking how to calculate the mass of aluminum nitride (AlN) needed to create a 0.2 M solution at a volume of 0.20 L. To solve this, we must first determine the molar mass of AlN. The molar mass of aluminum (Al) is 26.98 g/mol and that of nitrogen (N) is 14.01 g/mol, giving us a total molar mass for AlN of 26.98 g/mol + 14.01 g/mol = 40.99 g/mol.

Now, to make a 0.2 M solution, we require 0.2 moles of solute per liter of solution. Since the desired volume is 0.20 L, the moles of AlN needed are 0.2 M × 0.20 L = 0.04 moles. Using the molar mass of AlN, the mass required can be calculated as:

0.04 moles × 40.99 g/mol = 1.6396 g

Therefore, we round off to obtain the option closest to our calculation:

1.8 g of aluminum nitride is needed to prepare the solution, so the correct answer would be (b) 1.8 g.