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If 20.0 mLmL of a 0.100 NN acid solution is needed to reach the end point in titration of 32.5 mLmL of a base solution, what is the normality of the base solution

Sagot :

Answer:

The correct answer is "0.051 N".

Explanation:

Given

Volume,

[tex]V_1=20.0 \ mL[/tex]

[tex]V_2=32.5 \ mL[/tex]

Normality,

[tex]N_1=0.1 \ N[/tex]

[tex]N_2=?[/tex]

As we know,

⇒ [tex]N_1V_1=N_2V_2[/tex]

or,

⇒ [tex]N_2=\frac{N_1V_1}{V_2}[/tex]

On putting the values, we get

⇒      [tex]=\frac{0.1\times 20.0}{32.5}[/tex]

⇒      [tex]=\frac{2}{32.5}[/tex]

⇒      [tex]=0.051 \ N[/tex]

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