A chemist needs to strengthen a 15% alcoholic solution to one of 32% alcoholic. How much pure alcohol should be added to 800ml of 15% solution?
Answer
362.4k+ views
Hint: We have to change the concentration of alcohol of 800ml solution by adding ‘x’ ml of pure alcohol. We will use percentages concept for calculations.
Complete step-by-step answer:
From the given information, we have 800ml of solution with 15% alcohol.
The alcohol in 800ml solution = $800 \times \dfrac{{15}}{{100}} = 120ml$
Now we need to add some pure alcohol to this 800ml to change the alcohol percentage to 32%.
Let ‘x’ ml of pure alcohol (100% alcohol) is added.
Then our resulting solution = (800 + x) ml
Amount of alcohol in the resulting solution = (120 + x) ml
We want the alcohol percentage as 32%.
$ \Rightarrow \dfrac{{Amount\;of\;alcohol}}{{Total\;amount\;of\;solution}} \times 100 = 32$
$ \Rightarrow \dfrac{{x + 120}}{{800 + x}} \times 100 = 32$
On simplification of the above equation,
$\eqalign{
& \Rightarrow 100x + 12000 = 32x + 25600 \cr
& \Rightarrow 100x - 32x = 25600 - 12000 \cr
& \Rightarrow 68x = 13600 \cr} $
$ \Rightarrow x = \dfrac{{13600}}{{68}} = 200$
$\therefore $ The amount of pure alcohol to be added is 200ml.
Note: While solving these kinds of problems, we need to clearly understand the given information. We have to form algebraic equations for calculating the required values.
Complete step-by-step answer:
From the given information, we have 800ml of solution with 15% alcohol.
The alcohol in 800ml solution = $800 \times \dfrac{{15}}{{100}} = 120ml$
Now we need to add some pure alcohol to this 800ml to change the alcohol percentage to 32%.
Let ‘x’ ml of pure alcohol (100% alcohol) is added.
Then our resulting solution = (800 + x) ml
Amount of alcohol in the resulting solution = (120 + x) ml
We want the alcohol percentage as 32%.
$ \Rightarrow \dfrac{{Amount\;of\;alcohol}}{{Total\;amount\;of\;solution}} \times 100 = 32$
$ \Rightarrow \dfrac{{x + 120}}{{800 + x}} \times 100 = 32$
On simplification of the above equation,
$\eqalign{
& \Rightarrow 100x + 12000 = 32x + 25600 \cr
& \Rightarrow 100x - 32x = 25600 - 12000 \cr
& \Rightarrow 68x = 13600 \cr} $
$ \Rightarrow x = \dfrac{{13600}}{{68}} = 200$
$\therefore $ The amount of pure alcohol to be added is 200ml.
Note: While solving these kinds of problems, we need to clearly understand the given information. We have to form algebraic equations for calculating the required values.
Last updated date: 25th Sep 2023
•
Total views: 362.4k
•
Views today: 8.62k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

What is the basic unit of classification class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
