The answer is on the paper
Answer:
16 g/L
Explanation:
Given that:-
Mass of sucrose added = 1.2 g
Volume of water = 75 mL = 0.075 L ( 1mL = 0.001 L )
%w/v is defined as the mass of the solution in 1 L of the solution.
So,


Answer:
T°fussion of solution is -18°C
Explanation:
We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1
First of all, we apply boiling point elevation
ΔT = Kb . m . i
ΔT = Boiling T° of solution - Boiling T° of pure solvent
Kb = ebuliloscopic constant
105°C - 100° = 0.512 °C kg/mol . m . 1
5°C / 0.512 °C mol/kg = m
9.7 mol/kg = m
Now that we have the molality we can apply, the Freezing point depression.
ΔT = Kf . m . i
Kf = cryoscopic constant
0° - (T°fussion of solution) = 1.86 °C/m . 9.76 m . 1
- (1.86°C /m . 9.7 m) = T°fussion of solution
- 18°C = T°fussion of solution
20 g/cm3 I got this because 120 divided by 6 is 20
Answer:
(CH₃)₃COCH3₃ and (CH₃)₂CHOCH₂CH₃
Explanation:
Isomers are compounds which have the same molecular formula. Constitutional isomers have different connectivity; the atoms are connected in different ways.
1. (CH₃)₃COCH₃
2. (CH₃)₂CHOCH3₃
3. (CH₃)₂CHOCH₂CH₃
Molecules 1 and 3 have the same formula (C₅H₁₂O) and are isomers. Molecule 2 is not an isomer. From the structural formula, it is clear that Molecules 1 and 3 have different connectivity.