Answer:
- 0.0413°C ≅ - 0.041°C (nearest thousands).
Explanation:
- Adding solute to water causes the depression of the freezing point.
<em>ΔTf = Kf.m,</em>
Where,
ΔTf is the change in the freezing point.
Kf is the freezing point depression constant (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>Molality is the no. of moles of solute per kg of the solution.</em>
- <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>
<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>
∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.
<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>
We will use Boyle's law. P = 1/V which simply means pressure is inversely proportion to volume at a constant temperature. We will assume temperature here does not change. Since we have two pressure/volume points, we use this version of Boyle's law.
P1 V1 = P2 V2
P is pressure and V is volume
In our case:
P1 is 736, V1 is 4.2 while V2 is 9.0 but P2 is unknown. This is what we need to find out.
736 * 4.2 = P2 * 9.0
P2 = 736 * 4.2 / 9.0
P2 = 3091.2 / 9.0
P2 = 343.5
Therefore the new pressure 343 mm/hg
Here we see that the increase in volume causes decrease in pressure in total agreement with Boyle's law.
The correct answer to this question is the unique atomic number which would be B
Hey buddy I am here to help!
1. C
2. A
3. A & B
4. C
5. C
6. A
7. A
8. A & C
Hope it helps!
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