Question

How much pure alcohol must be added to 400 mL of a 15% solution to make its strength 32%.

Answer 1

Hint: In this question, the concept of percentages will be used. The relation between the percentage and the amount of substance in a mixture is given by-
$$\dfrac{\mathrm{Volume}\;\mathrm{of}\;\mathrm{dissolved}\;\mathrm{substance}\;}{\mathrm{Total}\;\mathrm{volume}\;\mathrm{of}\;\mathrm{liquid}}\times100=\mathrm{percentage}$$

Complete step-by-step solution -
Now, we have been given that initially 15% of alcohol is present in a 400 mL solution. We will first find the volume of alcohol in the initial solution using the formula. Let the amount of alcohol be v mL, then-
$\dfrac{\mathrm v}{400}\times100=15\\$
v = 60 mL
This is the initial amount of alcohol. Now, let us assume that x mL of alcohol is added to the solution. The new volume of alcohol will be 60+x and the new volume of solution will be 400+x. It is given that the strength of this solution is 32%. So, the equation formed will be-
$\dfrac{60+\mathrm x}{400+\mathrm x}\times100=32$
$\Rightarrow 60 + x = 0.32(400 + x)$
$\Rightarrow 60 + x = 128 + 0.32x $
$\Rightarrow 0.68x = 68 $
$\Rightarrow x = 100 mL $
Therefore, 100 mL of alcohol should be added to make the strength 32%. This is the required answer.

Note: In this question, no specific formula is needed, we just need to apply some logic and express the percentage in terms of volume. Also, writing the right units in the answer is a must.