You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution
1 answer:
<u>Answer:</u> The mass of solution having 768 mg of KCN is 426.66 grams.
<u>Explanation:</u>
We are given:
0.180 mass % of KCN solution.
0.180 %(m/m) KCN solution means that 0.180 grams of KCN is present in 100 gram of solution.
To calculate the mass of solution having 768 mg of KCN or 0.786 g of KCN (Conversion factor: 1 g = 1000 mg)
Using unitary method:
If 0.180 grams of KCN is present in 100 g of solution.
So, 0.768 grams of KCN will be present in = of solution.
Hence, the mass of solution having 768 mg of KCN is 426.66 grams.
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Answer:
the atomic packing factor of Sn is 0.24
Explanation:
a = b = 5.83A and c = 3.18A.
Volume of unit cell = a²c
= (5.83)² * 3.18 * 10⁻²⁴ cm³
= 1.08 * 10⁻²²cm³
Volume of atoms =
(∴ BCC, effective number of atom is 2)
Volume of atoms =
= 2.55*10⁻²³cm³
Answer:
Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
Use the molar mass as a ratio.
We want to convert 3.14 moles to grams, so we multiply by that value.
The units of moles of PI₃ cancel.
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
- 1292.698
The 2 in the ones place tells us to leave the 9.
3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>