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
Answer:
specialized if you add one more I cause I'm pretty sure there is supposed to be one more i.
Gamma rays have the highest penetrating power so it can only be stopped by thick layers of dense metal. :)
