Question

Question #3eb4a

Answer 1

`3 1/3` grams `-> 3.33bar3` grams

Explanation:

The most straight forward way to solve this is to use ratio.
Note that the shortcut method will be based on this approach.

Let the unknown weight of the substance be `x`

`"solute"/"solution" " "->" " ("mg")/("ml")" "->" "2/6 -=x/10`

To change 6 to 10 multiply by `10/6` so we have

`x/10= (2xx10/6)/(6xx10/6)`

Thus `x=2xx10/6 = 10/3 "grams"`