Respuesta :
Answer:
Concentration of original solution = 1.66
Explanation:
We know that
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
We have given concentration of NaOH = 0.1678
Volume of NaOH = 19.88 mL = 0.01988 L
So moles of NaOH = volume x concentration of NaOH
= [tex]0.01988\times 0.1678=0.00333mole[/tex]
Moles of [tex]H_2SO_4[/tex] in 10 mL of diluted solution = 1/2 x moles of NaOH
= [tex]\frac{1}{2}[/tex] x 0.00333 = 0.00166 mol
Moles of [tex]H_2SO_4[/tex] in 25 mL of original solution
= moles of H2SO4 in 250 mL of diluted solution
= [tex]\frac{250}{10}[/tex] x 0.00166 = 0.0415 mol
Concentration of original solution = [tex]\frac{moles}{volume}[/tex]
= [tex]\frac{0.0415}{0.025}=1.66[/tex]
