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 : The number of moles of
produced will be, 0.241 moles.
Solution : Given,
Mass of Fe = 27.0 g
Molar mass of
= 56 g/mole
First we have to calculate the moles of
.

Now we have to calculate the moles of 
The balanced chemical reaction is,

From the reaction, we conclude that
As, 4 mole of
react to give 2 mole of 
So, 0.482 moles of
react to give
moles of 
Thus, the number of moles of
produced will be, 0.241 moles.
Answer:
Explanation:B ice melting is a phiyscal change and the mass of the substance remains the same
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6