Answer : The molality of the solution is, 0.788 mole/Kg
Solution : Given,
Mass of solute, 
= 25 g
Mass of solvent, 
= 125 g
Molar mass of solute, = 253.8 g/mole
Molality : It is defined as the number of moles of solute present in one kilogram of solvent.
Formula used :
Now put all the given values in this formula, we get the molality of the solution.
Therefore, the molality of the solution is, 0.788 mole/Kg
Answer:
0.246 kg
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist adds 370.0mL of a 2.25 M iron(III) bromide (FeBr₃) solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>
<em />
We have 370.0 mL of 2.25 M iron(III) bromide (FeBr₃) solution. The moles of FeBr₃ are:
0.3700 L × 2.25 mol/L = 0.833 mol
The molar mass of iron(III) bromide is 295.56 g/mol. The mass corresponding to 0.833 moles is:
0.833 mol × 295.56 g/mol = 246 g
1 kilogram is equal to 1000 grams. Then,
246 g × (1 kg/1000 g) = 0.246 kg
The reaction is
CH3OH(l)→CH3OH(g)
To calculate ΔG°
ΔG°=ΔG°f of CH3OH(g)-ΔG°f of CH3OH(l)
=162.3kJ/mol - (-166.6kJ/mol
=4.3kJ/mol.
B) is an example of a Double Displacement Reaction:
<span>Ba Cl2 + K2 CO3 ---> Ba CO3 + 2K Cl
</span>A B C D A D B C
