Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
<span>Alkanes are unreactive except in combustion reactions.</span>
Total in pot=28 L
400 mL in each bowl
16 bowls filled
1000mL=1L
16 bowls(400mL/1 bowl)=6400mL
6400mL(1L/1000mL)=6.4L
28L-6.4L=21.6 L
CaCl2(aq) + K2CO3(aq) → 2 KCl(aq) + CaCO3(aq)
1.12 g
2.23 g
0.896 g
4.47 g
1.12 g
Hope it helps :)
You can use this formula to solve for density--> Density= PM/ RT, where P is pressure, M is molar mass, R is the gas constant and T is temperature.
P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k
density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L
