Answer : The molarity of solution is, 1.73 mole/L
Explanation :
The relation between the molarity, molality and the density of the solution is,
where,
![d=M[\frac{1}{m}+\frac{M_b}{1000}]](https://tex.z-dn.net/?f=d%3DM%5B%5Cfrac%7B1%7D%7Bm%7D%2B%5Cfrac%7BM_b%7D%7B1000%7D%5D)
d = density of solution = 
m = molality of solution = 2.41 mol/kg
M = molarity of solution = ?
= molar mass of solute (toluene) = 92 g/mole
Now put all the given values in the above formula, we get the molality of the solution.
![0.876g/ml=M\times [\frac{1}{2.41mol/kg}+\frac{92g/mole}{1000}]](https://tex.z-dn.net/?f=0.876g%2Fml%3DM%5Ctimes%20%5B%5Cfrac%7B1%7D%7B2.41mol%2Fkg%7D%2B%5Cfrac%7B92g%2Fmole%7D%7B1000%7D%5D)

Therefore, the molarity of solution is, 1.73 mole/L
Answer: The student should obtain <u>1.103 g of aluminum oxide </u>
Explanation:
- First we write down the equations that represent the aluminum hydroxide precipitation from the reaction between the aluminum nitrate and the sodium hydroxide:
Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3
Now, the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.
2Al(OH)3 → Al2O3 + 3H2O
- Second, we gather the information what we are going to use in our calculations.
Volumen of Al(NO3)3 = 40mL
Molar concentration of Al(NO3)3 = 0.541M
Molecular Weight Al2O3 = 101.96 g/mol
- Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

then we use the molar coefficients from the equations to discover the amount of Al2O3 moles produced

finally, we use the molecular weight of the Al2O3 to calculate the final mass produced.
