Question

A 30mL of alcohol solution is made of 60% water. How much of this solution must be replaced with a 90% alcohol solution so that the resulting solution has an alcohol content of 60%?​

Answer 1

Answer:

$Volume of solute= 50ml</p><p>Volume of solvent= 150ml</p><p>∴ Total volume= 150+50=200ml$

$Concentration= \frac{Volume of solute}{Volume of solution} \times \: 100$

$= \: \frac{50}{200} \times \: 100$

$= \: 25\%$

Answer 2

The quantity of % 60 water the solution replaced is 12 ml.

Step-by-step explanation:

Given:

30ml of % 60 water solution.

A part of %60 alcohol solution is replaced by%90 alcohol solution.

The % of alcohol in the resulting mixture is % 60.

To Find:

The quantity of % 60 water solution replaced.

Formula Used:

Quantity of %90 alcohol solution / Quantity of %40 alcohol solution

= (Strength of %90 alcohol solution -Mean Strength) /(Mean Strength- Strength of %40 alcohol solution )    ------------------- formula no.01

The above is per the rule of the allegation.

Solution:

As given- 30ml of % 60 water solution.        

               That means 30 ml of % 40 alcohol solution

As given - A part of % 60 water solution is replaced by%90 an alcohol solution.

 As given -The % of alcohol in the resulting mixture is % 60.

Strength of %40 alcohol solution =  %40

Strength of %90 alcohol solution =  %90

Mean Strength= %60

Applying formula no.01.

$\frac{Quantity \ of \% 90 \ alcohol \ solution}{Quantity \ of \% 40 \ alcohol \ solution} = \frac{(90-60)}{(60-40)}$

                                              $=\frac{30}{20} = \frac{3}{2}$

The quantity of % 40 alcohol solution replaced $= \frac{2}{5} \times 30\ ml$

                                                                                 $= \frac{60}{5} = 12\ ml$

That means  also the quantity of % 60 water solution replaced = 12 ml

Thus, the quantity of % 60 water solution replaced is 12 ml.