Question
A chemistry student needs 15.00g of bromobutane for an experiment. She has available 220.g of a 30.2% w/w solution of bromobutane in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.
Round your answer to 5 significant digits.
Answer:
49.669g
Explanation:
Given
220 g of solution has 30.2% w/w bromobutane
15.00g of bromobutane needed?
Required
Since the 220g solution has 30.2% w/w bromobutane
Then
The solution has 0.302 * 220
= 66.44g of bromobutane
From here. We can Calculate the expected grams required for 15.0g of bromobutane
, if 220 g of solution contains 66.44 g bromobutane,
The 15.0g will contain
=>
220 g solution = 66.44 g bromobutane
Y g solution = 15.00 g bromobutane
-----------Credit Multiply-----_-----
Y * 66.44 = 15 * 220
Y * 66.44 = 3300 ------- Divide both sides by 66.44
Y = 3300/66.44
Y = 49.66887g
Y = 49.669g --------- Approximated
The student should use 49.669g of the solution
