A569324 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.

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 :

Where = Initial concentration

= Initial volume

= Final concentration

= 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 .

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

is in 'mL' and is in 'L' , so we will first convert into 'L'

=

= 0.569324 L

Now we will plug in the values of , in dilution equation and will calculate the value of .

(Initial concentration) = 11.38 M

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"