Question

The sample calculation of sodium hydroxide molarity for trial #1 is _M.(4 sig figs, no units, no scientific notation)

Answer 1

Q1. sodium hydroxide molarity for trial #1 = 0.1006 M

Q2. average sodium hydroxide molarity = 0.1009 M

Explanation

For trial #1

mass KHP = 0.306 g

moles KHP = (mass KHP) / (molar mass KHP)

moles KHP = (0.306 g) / (204.22 g/mol)

moles KHP = 1.50 x 10-3 mol

moles NaOH = moles KHP

moles NaOH = 1.50 x 10-3 mol

volume NaOH used = (final reading) - (initial reading)

volume NaOH used = (14.9 mL) - (0.0 mL)

volume NaOH used = 14.9 mL

volume NaOH used = 0.0149 L

molarity of NaOH = (moles NaOH) / (volume NaOH used in Liters)

molarity of NaOH = (1.50 x 10-3 mol) / (0.0149 L)

molarity of NaOH = 0.1006 M

Similarly, for trial #2, molarity of NaOH = 0.1012 M

average molarity = (M1 + M2) / 2

average molarity = (0.1006 M + 0.1012 M) / 2

average molarity = 0.1009 M