Question

A 569324 mL NaCl solution is diluted to a volume of 1.33 L and a concentration of 4.87 M. What was the initial concentration? Give the molar concentration rounded to the hundredths place without units.

Answer 1

Since 569324 mL = 569.324 L which is much more than 1.33 L that means it cannot be diluted to a volume of 1.33 L, so this must be 569.324 mL.

This question is solved by using the dilution equation which is :
`M_(1)V_(1)=M_(2)V_(2)`

Where
`M_(1)` = Initial concentration

`V_(1)` = Initial volume

`M_(2)` = Final concentration

`V_(2)` = Final volume

In our question , the given information is : V1 = 569.324 mL

M2 = 4.87 M and V2 = 1.33 L and we need to find
`M_(1)`.

When we use dilution equation we need to make sure that both the units of volume have same units.

`V_(1)` is in 'mL' and
`V_(2)` is in 'L' , so we will first convert
`V_(1)` into 'L'

`V_(1)` =
`(569.324 mL)* ((1 L)/(1000 mL))`

`V_(1)` = 0.569324 L

Now we will plug in the values of
`V_(1),M_(2),V_(2)` , in dilution equation and will calculate the value of
`M_(1)`.

`M_(1)V_(1) = M_(2)V_(2)`

`M_(1)* 0.569324 L = 4.87 M* 1.33 L`

`M_(1) = ((4.87 M* 1.33 L))/(0.569324 L)`

`M_(1)` (Initial concentration) = 11.38 M