Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
<span>Most molecules containing carbon are organic. All organic compounds, from basic elements through complex living creatures, contain some degree of carbon, which is one of the most basic components of any organic molecule.</span>
mole ratios of hydrazine should be 1:2. I could be wrong. Are there any options to choose from?
1 mole Hg --------------- 6.02x10²³ atoms
?? moles Hg ------------ 1.30x10⁷ atoms
(1.30x10⁷) x 1 / 6.02x10²³ => 2.159x10⁻¹⁷ moles
