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A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.

(A) 0.270
(B) 1.48
(C) 0.675
(D) 270
(E) 675

Respuesta :

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = concentration of [tex]HNO_3[/tex] = 13.5 M

[tex]M_2[/tex] = concentration of diluted solution = ?

[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] = 25.0 ml  = 0.0250 L

conversion used : (1 L = 1000 mL)

[tex]V_2[/tex] = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

[tex]13.5M\times 0.0250L=M_2\times 0.500L[/tex]

[tex]M_2=0.675M[/tex]

Therefore, the concentration of the diluted solution is 0.675 M

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