what is the molarity of an HCI solution if 25.0 ml of 0.185 M NaOH is required to neutralize 0.0200 L of HCI?

0.231 M.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, the data obtained from the question:

Volume of base, NaOH (Vb) = 25 mL

Molarity of base, NaOH (Mb) = 0.185 M

Volume of acid, HCl (Va) = 0.0200 L

Molarity of acid, HCl (Ma) =?

Next, we shall convert 0.0200 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.0200 L = 0.0200 L × 1000 mL /1 L

0.0200 L = 20 mL

Thus, 0.0200 L is equivalent to 20 mL.

Finally, we shall determine the molarity of the acid solution as follow:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Volume of base, NaOH (Vb) = 25 mL

Molarity of base, NaOH (Mb) = 0.185 M

Volume of acid, HCl (Va) = 20 mL

Molarity of acid, HCl (Ma) =?

MaVa / MbVb = nA/nB

Ma × 20 / 0.185 × 25 = 1

Ma × 20 / 4.625 = 1

Cross multiply

Ma × 20 = 4.625 × 1

Ma × 20 = 4.625

Divide both side by 20

Ma = 4.625 / 20

Ma = 0.231 M

Thus, the molarity of the acid solution is 0.231 M.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be