ETHANOL (C2H5OH) HAS A DENSITY OF 0.80g/ml. IF A 10%(v/v) MIXTURE OF ETHANOL IN WATER WAS PREPARED HOW MANY GRAMS OF ETHANOL WOULD BE IN 100ML OF THE MIXTURE? (ASSUMING MIXED VOLUME IS THE SUM OF THE PARTIAL VOLUME)
PLEASE don't post in caps. All caps makes the post difficult to read.
10% v/v means 10 mL ethanol/100 mL soln.
So 10 mL = how many grams? mass = volume x density
To find out how many grams of ethanol are in 100 mL of a 10% ethanol mixture in water, you need to calculate it using the density and volume percent information.
Here's how you can do it step by step:
Step 1: Convert the volume percent to a decimal fraction.
A 10% (v/v) mixture means that for every 100 mL of the mixture, 10 mL is ethanol. So, 10% can be written as 0.1 (10/100 = 0.1).
Step 2: Calculate the volume of ethanol in the 100 mL mixture.
Since 10% of the mixture is ethanol and the total volume is 100 mL, the volume of ethanol in the mixture is 0.1 x 100 mL = 10 mL.
Step 3: Convert the volume of ethanol to grams.
To do this, you need to use the density of ethanol.
Density = Mass / Volume
Given that the density of ethanol is 0.80 g/mL, you can rearrange the formula to solve for mass (m):
Mass = Density x Volume
Mass = 0.80 g/mL x 10 mL = 8 grams
Therefore, there would be 8 grams of ethanol in 100 mL of the mixture.