The molecular weight of hemoglobin can be calculated using osmotic pressure
Osmotic pressure is a colligative property and it depends on molarity as
πV = nRT
where
π = osmotic pressure
V = volume = 1mL = 0.001 L
n = moles
R = gas constant = 0.0821 L atm / mol K
T = temperature = 25°C = 25 + 273 K = 298 K
Putting values we will get value of moles
we know that
Therefore
Answer:
a) 1.248 x 10⁷ kg
b) 1.248 x 10⁴ Mg
c) 1.248 x 10¹³ mg
d) 1.248 x 10⁴ ton
Explanation:
a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):
1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg
b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):
1.248 x 10¹⁰ g * ( 1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg
c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):
1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg
d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):
1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton
The answer to this question is 159.609 g/mol
Answer:
V = 27.98 L
Explanation:
Given data:
Mass of CO₂ = 33.0 g
Pressure = 500 torr
Temperature = 27°C
Volume occupied = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 33.0 g/ 44 g/mol
Number of moles = 0.75 mol
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
Now we will convert the temperature.
27+273 = 300 K
Pressure = 500 /760 = 0.66 atm
By putting values,
0.66 atm×V = 0.75 mol × 0.0821 atm.L/ mol.K × 300 K
V = 18.47 atm.L/0.66 atm
V = 27.98 L
