how many grams of ethyl alcohol are needed to produce 1 L of a 2 mM solution if 1 mole =46g of ethyl alcohol. how do you calculate?
molarity = moles/L
So you want 2 millimoles.
mols = grams/molar mass
0.002 = g/46
g = 0.002 x 46 = ?? grams.
Add ?? grams to a 1 L volumetric flask and make to the mark.
To calculate the number of grams of ethyl alcohol needed to produce a 2 mM solution in 1 L, you will need to follow these steps:
Step 1: Determine the molar mass of ethyl alcohol.
The molar mass of ethyl alcohol is given as 46 g/mole.
Step 2: Convert mM to moles.
1 mM means 1 millimole per liter (1 mmol/L). Since we want to find the number of moles in 1 L, there would be 2 mmol in 1 L.
Step 3: Convert moles to grams.
Using the molar mass of ethyl alcohol (46 g/mole), we can calculate the grams of ethyl alcohol required.
Calculation:
Number of moles = 2 mmol = 2/1000 moles.
Number of grams = Number of moles × Molar mass
= (2/1000) moles × 46 g/mole
= 0.092 grams.
Therefore, approximately 0.092 grams of ethyl alcohol are needed to produce 1 L of a 2 mM solution.
To calculate the number of grams of ethyl alcohol needed to produce a 2 mM solution, we first need to understand what a 2 mM solution means.
"Molarity" (M) is a unit of concentration that measures the number of moles of solute in a liter of solution. In this case, 2 mM means that there are 2 millimoles (mmol) of ethyl alcohol in one liter of solution.
To determine the number of moles of ethyl alcohol needed, we need to convert the given millimoles to moles:
2 mmol = 2 * 10^(-3) mol (since 1 mmol = 10^(-3) mol)
Next, we can use the molar mass of ethyl alcohol to convert moles to grams. Given that 1 mole of ethyl alcohol equals 46 grams, we can set up the following equation:
2 * 10^(-3) mol ethyl alcohol * 46 g/mol ethyl alcohol = grams of ethyl alcohol needed
Simplifying the equation:
(2 * 10^(-3) * 46) g = grams of ethyl alcohol needed
Calculating the result:
(2 * 10^(-3) * 46) g = 0.092 g
Therefore, you would need approximately 0.092 grams of ethyl alcohol to produce 1 liter of a 2 mM solution.