The volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.
<h3>How to calculate volume?</h3>
The volume of a given mass of gas can be calculated using the following formula:
PV = nRT
Where;
- P = pressure
- V = volume
- R = gas law constant
- T = temperature
- n = number of moles
According to this question, 0.98 moles of oxygen gas at 275 k contains a pressure of 2.0 atm. The volume is calculated as follows:
2 × V = 0.98 × 0.0821 × 275
2V = 22.13
V = 11.06L
Therefore, the volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.
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1 mole of nitric acid produce 1 mole of ammonium nitrate.
moles in 5000 kg of ammonium nitrate :
( molecular mass of ammonium nitrate is 80 gm/mol )
So, number of moles of nitric acid required are also 62500 moles.
Mass of 62500 moles of nitric acid :

Hence, this is the required solution.
Ionic molecules have higher boiling point than covalent molecules.
So that leaves Na2SO4 and NaCl.
The ionic molecule with higher charge will have higher boiling point.
Na2SO4 have ions with charge +-2.
NaCl have ions with charge +-1.
So the answer is C.
1)
<span>m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = </span><span>58.5 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
</span>V(H2O) = 250ml = 250g = 0.25 kg<span>
</span><span>molality of NaCl:
</span><span>
n(NaCl)=m/M=1.95/58.5= 0.033 mole
</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation
Since NaCl dissociates into 2 ions in solution:
<span>
</span>milliosmolality of NaOH = 132 x 2 = 264 osmol<span>es/kg
</span>
2)
m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
V(H2O) = 250ml = 250g = 0.25 kg
molality of glucose:
n(gl)=m/M=9/180= 0.05 mole
molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg
milimolality of glucose = 0.132/0,001 = 200 mmole/kg
milliosmolality of glucose = milimolality x N of ions formed in dissociation
Since glucose does not dissociate, milimolality and milliosmolality are same:
milliosmolality of glucose = 200 osmoles/kg
3)
The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.
In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.