Answer:
The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Explanation:
We are given that
Aqueous solution that contains 22.9% NaOH by mass means
22.9 g NaOH in 100 g solution.
Mass of NaOH(WB)=22.9 g
Mass of water =100-22.9=77.1
Na=23
O=16
H=1.01
Molar mass of NaOH(MB)=23+16+1.01=40.01
Number of moles =
Using the formula
Number of moles of NaOH
Molar mass of water=16+2(1.01)=18.02g
Number of moles of water
Now, mole fraction of NaOH
=
=0.882
Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Answer:
S = 21.92 %
F = 78.08 %
Explanation:
To find the percent composition of each element in SF6, we must find the molar mass of SF6 first.
Molar mass of SF6 = 32 + 19(6)
= 32 + 114
= 146g/mol
mass of Sulphur (S) in SF6 = 32g
mass of Fluorine (F) in SF6 = 114g
Percent composition = mass of element/molar mass of compound × 100
- % composition of S = 32/146 × 100 = 21.92%.
- % composition of F = 114/146 × 100 = 78.08%.
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