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nexus9112 [7]
3 years ago
12

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
Chemistry
1 answer:
salantis [7]3 years ago
8 0

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

Explanation :

Using neutralization law,

M_1V_1=M_2V_2

where,

M_1 = concentration of HNO_3 = 13.5 M

M_2 = concentration of diluted solution = ?

V_1 = volume of HNO_3 = 25.0 ml  = 0.0250 L

conversion used : (1 L = 1000 mL)

V_2 = 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.

13.5M\times 0.0250L=M_2\times 0.500L

M_2=0.675M

Therefore, the concentration of the diluted solution is 0.675 M

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\boxed{\text{2.00 mol}}

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We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

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