Answer:
3.50 molal
Explanation:
Molality → Moles of solute / kg of solvent.
Let's convert the solvent's mass from g to kg
16.2 g . 1kg / 1000 g = 0.0162 kg
Let's determine the moles from the solute
2.61 g . 1 mol / 46 g = 0.0567 moles
Molality → 0.0567 mol / 0.0162 kg = 3.50 m
First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48
Explanation:
so for this u have to use this equation where
Moles = number of particle/6.02×10^23
= 3.045 × 10^24/6.02×10^23
= 5.0581
write it to 3 S.F so 5.06 moles
