
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
549.6k+ 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.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE

The southernmost point of the Indian mainland is known class 7 social studies CBSE

How many crores make 10 million class 7 maths CBSE

List of coprime numbers from 1 to 100 class 7 maths CBSE

One lakh eight thousand how can we write it in num class 7 maths CBSE

AIM To prepare stained temporary mount of onion peel class 7 biology CBSE
