The osmotic pressure of a solution is a colligative property, which means that it depends on the number of particles of solute in the solution.
Formula: Osmotic pressure = MRT, where M is the molarity of the solution, R is the universal constant of ideal gases and T is the absolute temperature of the solution.
So, the answer is the option .: the osmotic pressure of a solution increases as the number of particles of solute in the solution increases.
Hey there:
a) atomic mass:
Carbon =<span>12.0107 g/mol
</span>Hydrogen = <span>1.00794 g/mol
Oxygen = </span><span>15.9994 g/mol
</span>
Therefore:
C12H22O11 =
12 * 12.0107 + 1 * 1.00794 + 16 * 15.9994 => <span>342.29648 g/mol
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b) number of moles:
n = m / mm
n = 3.115 / </span><span>342.29648
n = 0.0091 moles
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hope this helps!</span>
Answer:
The correct option is;
d 4400
Explanation:
The given parameters are;
The mass of the ice = 55 g
The Heat of Fusion = 80 cal/g
The Heat of Vaporization = 540 cal/g
The specific heat capacity of water = 1 cal/g
The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice
The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal
The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change
The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal
The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice
The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal
The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal
However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.
The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal
The temperature,internal energy,and kinetic energy will all increase
Answer:
4.285 L of water must be added.
Explanation:
Hello there!
In this case, for this dilution-like problems, we need to figure out the final volume of the resulting solution so that we would be able to obtain the correct volume of diluent (water) to be added. In such a way, we can obtain the final volume, V2, as shown below:
Thus, by plugging in the initial molarity, initial volume and final molarity (0.587 M) we obtain:
It means we need to add:
Of diluent water.
Regards!
