
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
610.2k+ 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
You are awaiting your class 10th results Meanwhile class 7 english CBSE

Questions & Answers - Ask your doubts

A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Two Planoconcave lenses 1 and 2 of glass of refractive class 12 physics CBSE

Trending doubts
Convert 200 Million dollars in rupees class 7 maths CBSE

Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE

What are the controls affecting the climate of Ind class 7 social science CBSE

Write a letter to the editor of the national daily class 7 english CBSE

Welcome speech for Christmas day celebration class 7 english CBSE

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


