Question

If 60 ml of water contains 12% of chlorine, how much water must be added in order to create a 8% chlorine solution?

Answer 1

ɢɪᴠᴇɴ :

• total quantity of water is 60 ml
• contains 12% of chlorine

ᴛᴏ ғɪɴᴅ:

• the quantity of water must be added in order to create a 8% chlorine

sᴏʟᴜᴛɪᴏɴ :

Let x ml of chlorine be present in water.

Then, $\sf{\frac{12}{100} = \frac{x}{60}}$

→ x = $\sf{\frac{12}{100} × 60}$

→ x = 7.2 ml

Therefore, 7.2 ml is present in 60 ml of water.

In order for this 7.2 ml to constitute 8% of the solution, we need to add extra water. Let this be y ml.

Then, $\sf{\frac{8}{100} = \frac{7.2}{y}}$

→ y = 90 ml.

So in order to get a 8% chlorine solution, we need to add 90 - 60 = 30 ml of water.

Answer 2

Concept:

Two units are used to measure objects: mass and volume. Volume is the amount of space an item occupies, whereas mass is the quantity of substance it contains.

Given:

The volume of water = 60 mL

% of chlorine = 12%

% of chlorine solution = 8%

Find:

If 60 ml of water contains 12% of chlorine, how much water must be added in order to create an 8% chlorine solution?

Solution:

Let x mL of chlorine be present in the water.

The mass of chlorine in 12% chlorine solution in 60 mL water = 12/100 * 60 mL

= 0.12 * 60 mL

= 7.2 mL

Thus, 7.2 g of chlorine is present in 60 mL of water.

For the 8% chlorine solution, we have to add more water. So,

8 / 100 = 7.2 mL / Volume of water

0.08 = 7.2 mL / Volume of water

The volume of water = 7.2 mL / 0.08

= (720 / 8) mL

= 90 mL

Hence, if 60 mL of water contains 12% of chlorine, then 90 mL of water must be added in order to create an 8% chlorine solution.

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