how many grams of aluminum are in 25.0 ml of a 14%(w/v) solution of Al2O3?
14g Al2O3 w/v means 14g Al2O3/100 ml soln.
14 g Al2O3 x (2*Al/Al2O3 = 14 x (2*27/102) = 7.4 g Al which means 7.4 g Al/100 mL soln. So how much would be in 25.0 mL? Something like 1/4 that? I've estimated my numbers so you need to go through and recalculate everything.
i am not clear? i came up with .074g??
To determine the number of grams of aluminum in a 14% (w/v) solution of Al2O3, you need to consider the relationship between volume, concentration, and the molar mass of Al2O3.
Here's a step-by-step explanation of how to calculate it:
Step 1: Understand the 14% (w/v) concentration:
The 14% (w/v) concentration means that 14 grams of Al2O3 are dissolved in 100 ml of solution. This can be written as 14 g/100 ml = 0.14 g/ml.
Step 2: Calculate the volume of the solution in grams:
You are given the volume as 25.0 ml. To convert it to grams, multiply it by the density of the solution. Assuming the density is 1 g/ml for simplicity, the volume in grams is 25.0 g.
Step 3: Calculate the amount of Al2O3 in grams:
Since the concentration is given in grams per milliliter, multiply the volume (25.0 ml) by the concentration (0.14 g/ml):
25.0 ml x 0.14 g/ml = 3.5 grams of Al2O3.
Step 4: Calculate the amount of aluminum:
Since Al2O3 has a molar mass of 101.96 g/mol and consists of 2 aluminum atoms, you need to calculate the amount of aluminum in 3.5 grams of Al2O3.
The molar mass of aluminum is 26.98 g/mol, and there are 2 aluminum atoms in Al2O3.
So, the amount of aluminum can be calculated as:
(3.5 g Al2O3) x (2 mol Al / 101.96 g Al2O3) x (26.98 g Al / 1 mol Al) = 1.327 grams of aluminum.
Therefore, there are approximately 1.327 grams of aluminum in 25.0 ml of a 14% (w/v) solution of Al2O3.