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'

$V_{1}$ = $(569.324 mL)\times (\frac{1 L}{1000 mL})$

$V_{1}$ = 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 $M_{1}$.

$M_{1}V_{1} = M_{2}V_{2}$

$M_{1}\times 0.569324 L = 4.87 M\times 1.33 L$

$M_{1} = \frac{(4.87 M\times 1.33 L)}{0.569324 L}$

$M_{1}$ (Initial concentration) = 11.38 M