Chemistry question - Normality?

Basic techniques in normality calculation:
1. No. of equivalents of the acid = No. of equivalents of the alkali (*needless to consider the mole ratio).
2. No. of equivalents = Normality × (Volume in L)

Solution to the question :
No. of equivalents of the alkali = (0.1879 eq/L) × (48.61/1000 mL) = 0.009134 eq
No. of equivalents of the acid = 0.009134 eq
normality of the acid = (0.009134 eq) / (50/1000 L) = 0.1827 eq/L = 0.1827 N

The answer: (C) 0.1827 N

For a solution, the number of moles of the hydronium ions must be equal to the number of moles of the hydroxide ions. A one normal solution contains one mole of these ions per liter of the solution. To determine the number of moles multiply the volume in liters by the normality.

For the hydroxide ion, n = 0.04861 * 0.1879 = 0.009133819
0.050 * N = 0.009133819
N = 0.009133819 ÷ 0.050
This is approximately 0.1827. C is the correct answer.

V1 * N1 = V2 * N2
If you use the equation above, you will get the same answer.
48.61 * 0.1879 = 50 * N
N = 9.133819 ÷ 50
I hope this is helpful for you.

(48.61 mL) x (0.1879 N) / (50 mL) = 0.1826 N
[Although four significant digits are not justified when given "50" mL.]