A scientist has 400ml of a 8% solution. how much water needs to be added to lower it to a 2% solution?

1200 ml  of water needs to be added to lower it to a 2% solution.

Step-by-step explanation:

Given that,

A scientist has 400ml of a 8% solution.

you need to find how water is required to lower the 400 ml to 2% solution.

Let 'x' be the amount of water added to lower it 2% solution.

The formula is given by,

volume of the liquid × % of solution = volume of the new liquid × new %

This can be denoted in chemical relationship ⇒ c₁v₁=c₂v₂ where c is concentration and v is volume .

Now, to find the total volume of new liquid in ml :

400 ml = v₁

0.08= c₁  

0.02=c₂

v₂=?

400 × 0.08 =v₂ × 0.02

v₂=1600 ml

To find the x value :

Amount to be added = volume of new liquid - volume of liquid before.

⇒ x =1600-400

⇒ x =1200 ml

∴ 1200 ml of water needs to be added to lower it to a 2% solution.

Let us assume x is the increase percentage in decimals, then we need to multiply original value with 1+x.

Similarly if we assume x is the decrease percentage in decimals, then we need to multiply original value with 1-x

A) Given that sam multiplied original value with 1.07 while calculating percentage.

Since 1.07>1, that is an increase.

And percentage change = x*100 = (1.07-1) * 100 = 7 %

B)Given that sam multiplied original value with 0.7 while calculating percentage.

Since 0.7<1, that is an decrease

And decrease percentage = x*100 = (1-0.7)*100 = 30%