What is the total charge on all of the electrons in two liters, 2 kg, of water? One mole of water has a mass of 18 g and each molecule of water contains 10 electrons.
Way I worked problem:
(2kg)(1000g/1kg)(1mol/18g)(10e-/1mol)(-1.60e-19/1e-)
What I did do wrong?
You need to know that there are 6.022 *10^23 molecules in 1 mole
so
there are (10 e- /1MOLECULE) (6*10^23 molecules/mol)
in other words your answer needs to be multiplied by Avagadro's number
I believe you have omitted the number of molecules in a mole; i.w., 6.02 x 10^23 molecules H2O/mol. Check my thinking.
-1.07e8 C
To calculate the total charge on all the electrons in two liters (2 kg) of water, you need to apply the correct conversion factors and units. Let's analyze the calculation you performed step by step:
(2kg) - This is the given mass of the water in kilograms, which is correct.
(1000g/1kg) - This is the conversion factor to convert kilograms to grams, which is also correct.
(1mol/18g) - This is the molar mass of water (H2O), which is correct. It represents the conversion factor to convert grams of water to moles of water.
(10e-/1mol) - This is the given information that there are 10 electrons per molecule of water, which is correct.
(-1.60e-19/1e-) - This is the charge of each individual electron, which is correct.
However, the mistake lies in the order of the conversion factors. When performing dimensional analysis, it is essential to arrange the units in the correct order so that the desired units cancel out, leaving only the desired unit in the numerator. The correct way to arrange the conversion factors is as follows:
(2kg) x (1000g/1kg) x (1mol/18g) x (10e-/1mol) x (-1.60e-19/1e-)
Now, let's simplify this calculation:
= (2kg) x (1000g/1kg) x (1mol/18g) x (10e-/1mol) x (-1.60e-19/1e-)
= (2 x 1000 x 10 x -1.60) x (kg x g x mol x e-) / (kg x g x mol x e-)
= -32,000 C
So, the correct calculation would result in a total charge of -32,000 Coulombs (C) on all the electrons in two liters (2 kg) of water.