Molarity = moles of solute/volume of solution in liters.
From this relation, we can figure out the number of moles of solute by multiplying the molarity of the solution by the volume in liters.
We have 53.1 mL, or 0.0531 L, of a 12.5 M, or 12.5 mol/L, solution. Multiplying 12.5 mol/L by 0.0531 L, we obtain 0.664 moles. So, in this volume of solution, there are 0.664 moles of solute (HCl).
Answer: 568g/mol
Explanation:
It should be noted that there are 40 atoms of carbon in lycopene.
Since mass of 1 carbon = 12g/mol
Mass of 40 carbon atoms = 40 × 12g/mol = 480g/mol
Let the molar mass of lycopene be represented by x.
Therefore the molar mass of carbon = x × mass percent of carbon in lycopene
x × 84.49% = 480g/mol
x × 0.8449 = 480g/mol
x = 480/0.8449
x = 568g/mol
The molar mass of lycopene is 568g/mol
Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
For this item, we need to assume that air behaves like that of an ideal gas. Ideal gases follow the ideal gas law which can be written as follow,
PV = nRT
where P is the pressure,
V is the volume,
n is the number of mols,
R is the universal gas constant, and
T is temperature
In this item, we are to determine first the number of moles, n. We derive the equation,
n = PV /RT
Substitute the given values,
n = (1 atm)(5 x 10³ L) / (0.0821 L.atm/mol.K)(0 + 273.15)
n = 223.08 mols
From the given molar mass, we calculate for the mass of air.
m = (223.08 mols)(28.98 g/mol) = 6464.9 g
<em>ANSWER: 6464.9 g</em>
