Respuesta :
Answer:
The answer is "256.33 ml".
Explanation:
Given
In [tex]25\%[/tex] solution means [tex]25 \ g \ Na_2CO_3 \ in\ 100\ ml[/tex] solution
[tex]\therefore[/tex] for [tex]675\ ml[/tex] solution mass of [tex]Na_2CO_3[/tex] needed is calculated as
[tex]\to 675 \ ml \times \frac{ 25 \g}{100\ ml} = 168.75\ g \ Na_2CO_3[/tex]
now calculating moles of the [tex]168.75\ g \ Na_2CO_3[/tex]
[tex]moles =\frac{mass}{molar\ mass}[/tex]
[tex]moles \ of \ Na_2CO_3 = \frac{168.75\ g}{105.9888 \ g \ per\ mol}[/tex]
[tex]= 1.592 \ mol \ Na_2CO_3[/tex]
by using the moles and molarity of the stock solution we calculate the volume of the stock solution:
molarity = moles / volume in liter
so
[tex]\text{volume in liter} = \frac{moles}{ molarity} = \frac{1.592\ mol}{ 6 \ mol\ per\ liter} = 0.26533\ L\\[/tex]
convert liter to ml [tex]\frac{( 0.26533\ L \times 1000 \ ml)}{1\ L} = 256.33\ ml[/tex]
