The percentage of oxygen by mass in water is greatest when
1 answer:
You might be interested in
Answer:
Option D.
Explanation:
- 6H₂ + P₄ → 4PH₃
First we convert the given reactant masses into moles, using their respective molar masses:
- 4.00 g H₂ ÷ 2 g/mol = 2 mol H₂
- 6.20 g P₄ ÷ 124 g/mol = 0.05 mol P₄
0.05 moles of P₄ would react completely with (6*0.05) 0.3 moles of H₂. There are more H₂ moles than required, meaning H₂ is in excess and P₄ is the limiting reactant.
Now we<u> calculate how many PH₃ moles could be formed</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.05 mol P₄ *
= 0.2 mol PH₃
Finally we <u>convert 0.2 mol PH₃ into grams</u>, using its <em>molar mass</em>:
- 0.2 mol PH₃ * 34 g/mol = 6.8 g
So the correct answer is option D.
Explanation:
Let the volume of the solution be 100 ml.
As the volume of glycol = 50 = volume of water
Hence, the number of moles of glycol =
=
=
= 0.894 mol
Hence, number of moles of water =
= 2.77
As glycol is dissolved in water.
So, the molality =
= 17.9
Therefore, the expected freezing point =
=
Thus, we can conclude that the expected freezing point is .