Gasoline sold in Ontario must contain at least 5.00% v/v ethanol? How much gasoline contains 1.50 L of ethanol?
my answer is 22.5 ml
is this correct
I don't think so but try it and see. Most answers you can check yourself.
(L solute/L solution)*100 = %
(1.50/22.5)*100 = 6.67% which isn't 5% so 22.5 mL must not be right.
To determine how much gasoline contains 1.50 L of ethanol, we need to calculate it based on the given minimum requirement of 5.00% v/v ethanol in gasoline sold in Ontario.
First, we need to calculate what 5.00% of the total volume of gasoline is. To do this, we'll convert the 5.00% to decimal form:
5.00% = 5.00 / 100 = 0.05
Next, we'll calculate the volume of gasoline that contains 1.50 L of ethanol by dividing the volume of ethanol by the percentage of ethanol in gasoline:
Volume of gasoline = Volume of ethanol / Percentage of ethanol
Volume of gasoline = 1.50 L / 0.05
Volume of gasoline = 30 L
Therefore, 1.50 L of ethanol would be present in 30 L of gasoline.
To convert this volume to milliliters (ml), we'll multiply the volume of gasoline by 1000 (since there are 1000 ml in 1 L):
Volume of gasoline in ml = 30 L * 1000 ml/L
Volume of gasoline in ml = 30,000 ml
So, 1.50 L of ethanol would be present in 30,000 ml (or 30 L) of gasoline, not 22.5 ml.
Therefore, your answer of 22.5 ml is incorrect. The correct answer is 30,000 ml.