Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
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The answer is (2) CO and NH3. The chemical reaction can only break down chemical bonds between atoms inside the molecule. The He and Xe is consists of one atom only, so they can not be broken down by chemical means.
The barium sulfate formed is 1.57 grams
The chemical reaction as
BaCl2 + Na2SO4 → 2NaCl + BaS04
The given date is
35 ml of 0.160 barium chloride
58 ml of 0.065 m sodium sulfate
35 ml of barium chloride = 0.0350 liters
58 ml of sodium sulfate = 0.050 liters
moles of barium chloride = 0.0160 / 0.0350
= 0.4571 mol
moles of sodium sulfate = 0.065 / 0.058
= 1.120
Barium sulfate = moles of barium chloride + sodium sulfate
= 0.4571 + 1.12
= 1.57 gram
Hence the gram of barium sulfate formed is 1.57 gram
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