Mol h2so4=0,6x3=1,8mol
MolOH-=Mol H+=2molh2so4=1,8x2=3,6mol
=> Mol NaOH=3,6mol=>V naoh= 3,6:4=0,9liters.
So, A is the correct answer
Answer: The correct answer is Option A.
Explanation:
To calculate the volume of NaOH, we use the equation used for the neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] = n-factor, molarity and volume of acid which is sulfuric acid.
[tex]n_2,M_2\text{ and }V_2[/tex] = n-factor, molarity and volume of base which is sodium hydroxide.
We are given:
[tex]n_1=2\\M_1=3M\\V_1=0.60L\\n_2=1\\M_2=4.0M\\V_2=?L[/tex]
Putting values in above equation, we get:
[tex]2\times 3\times 0.6=1\times 4\times V_2\\\\V_2=0.9L[/tex]
Hence, the correct answer is Option A.
