Final answer:
The Normality (N) of a 5.85 g KBr solution in 125 cm³ of water is calculated to be approximately 0.3933 N. None of the provided options match this calculated value. The closest provided option is c. 0.234 N, even though it is not accurate.
Step-by-step explanation:
The question is asking for the Normality (N) of a solution composed of 5.85 g of potassium bromide (KBr) dissolved in 125 cm³ of water. To find the normality, we need to know the number of equivalents of solute per liter of solution. Since KBr is an ionic compound that dissociates into one equivalent of K+ and one equivalent of Br-, the number of equivalents is the same as the number of moles for this compound.
The molar mass of KBr is 39.1 (for K) + 79.9 (for Br) = 119 g/mol. Thus, 5.85 g of KBr is 5.85 g / 119 g/mol = 0.04916 mol. Now, to determine normality, we divide the moles of equivalents by the volume of the solution in liters. Since we have 0.125 L (125 cm³), the normality is 0.04916 mol / 0.125 L = 0.3933 N.
However, none of the options given (a. 2.34 N, b. 0.936 N, c. 0.234 N, d. 4.68 N) match the calculated value. It's possible there may be a mistake in the question or the provided options. If an approximate answer must be chosen from the options given, option c (0.234 N) is the closest to the calculated normality, although it is not an accurate representation of the calculation.