I don’t really have a clear answer for you but this might help
The answer is 42 grams of NaF are there in 1 mole.
<h3>
What is a mole ?</h3>
A mole is defined as 6.022 × 10²³ atoms, molecules, ions, or other chemical units.
and the molar mass of a substance is defined as the mass of 1 mole of that substance, expressed in grams per mole.
It is equal to the mass of 6.022 × 10²³ atoms, molecules, or formula units of that substance.
Molar Mass , i.e. mass of 1 mole of NaF is sum of molar mass of Na and F
23 + 19
42 grams
Therefore 42 grams of NaF Sodium-fluoride are there in 1 mole.
To know more about mole
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Answer:
11.4
Explanation:
Step 1: Given data
- Concentration of the base (Cb): 0.300 M
- Basic dissociation constant (Kb): 1.8 × 10⁻⁵
Step 2: Write the dissociation equation
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
Step 3: Calculate the concentration of OH⁻
We will use the following expression.
![[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%3D%5Csqrt%7BKb%20%5Ctimes%20Cb%20%7D%20%3D%20%5Csqrt%7B1.8%20%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.300%20%7D%20%3D%202.3%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 4: Calculate the pOH
We will use the following expression.
![pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6](https://tex.z-dn.net/?f=pOH%20%3D-log%5BOH%5E%7B-%7D%20%5D%3D%20-log%282.3%20%5Ctimes%2010%5E%7B-3%7D%20M%29%20%3D%202.6)
Step 5: Calculate the pH
We will use the following expression.
<span>The density of the solution =1.05 g/ml.
</span><span>The total mass of the resulting solution is = 398.7 g (CaCl2 + water)
</span>
Find moles of CaCl2 and water.
Molar mass of CaCl2 = 110 (approx.)
Moles of CaCl2 = 23.7 / 110 = 0.22
so, moles of Cl- ion = 2 x 0.22 = 0.44 (because each molecule of CaCl2 will give two Cl- ions)
Moles of water = 375 / 18 = 20.83
Now, Mole fraction of CaCl2 = (moles of CaCl2) / (total moles)
total moles = moles of Cl- ions + moles of Ca2+ ions + moles of water
= 0.44 + 0.22 + 20.83
=21.49
So, mole fraction = 0.44 / (21.49) = 0.02
Guess what !!! density is not used. No need
