A student used 10 mL water instead of 30 mL for extraction of salt from mixture. How may this change the percentage of NaCl extracted?

It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:

360 g : 1,000 mL = x g : 30 mL

Now, crossing the products:

x · 1,000 mL = 360 g · 30 mL

x · 1,000 mL = 10,800 g mL

x = 10,800 g : 1,000

x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:

360 g : 1,000 mL = x g : 10 mL

Now, crossing the products:

x · 1,000 mL = 360 g · 10 mL

x · 1,000 mL = 3,600 g mL

x = 3,600 g : 1,000

x = 3.6 g

So, from 10 mL mixture, 3.6 g of NaCl could be extracted.

Now, let's compare:

If from 30 mL mixture, 10.8 g of NaCl could be extracted and from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:

3.6/10.8 = 1/3

Therefore, i t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.