Question

) Calculate:
Molality of a solution in which 0.52 moles of solute are dissolved in 500 g of water
) Molality of a solution in which 36 g of glucose is dissolved in 250 g of water.
( Weight
a) Molality of 90% (volume 4,80, (sp. gravity = 1.98 g/mL)
-) Calculate the percentage of aluminium in Al,0, (Atomic wt. of AI = 27, 0 = 16 )​

Answer 1

Explanation:

The molality of a solution essentially relates the number of moles of solute and the mass of the solvent.
More specifically, molality is defined as the number of moles of solute, which in your case is sodium chloride, NaCl, present in one kilogram of solvent, which is water.
To find the number of moles of sodium chloride present in your sample, sue the compound's molar mass

1kg × 0.52 moles/0.5 kg water = 1.04 mol kg^-1
by this prosses you can got all answer

Answer 2

(a)

Given:

Moles of solute $\[ = 0.52\,moles\]$

Mass of water $\[ = 500\,g\]$

To Find: Molality of the solution

Solution:

The molality of a solution can be given as:

$\[{\text{Molality = }}\frac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Weight}}\,{\text{of}}\,{\text{solvent}}\,{\text{(in}}\,{\text{kg)}}}}\]$

Therefore, the molality of the given solution is given as:

$\[\begin{gathered} {\text{Molality = }}\frac{{{\text{0}}{\text{.52}}}}{{{\text{0}}{\text{.5}}}} \hfill \\ {\text{Molality = 1}}{\text{.04}}\, \hfill \\ \end{gathered} \]$

Hence, the molality of the given solution is $1.04$.

(b)

Given:

Mass of solute $\[ = 36\,g\]$

Mass of water $\[ = 250\,g\]$

To Find: Molality of the solution

Solution:

The molality of a solution can be given as:

$\[{\text{Molality = }}\frac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Weight}}\,{\text{of}}\,{\text{solvent}}\,{\text{(in}}\,{\text{kg)}}}}\]$

Number of moles can be given as:

$\[{\text{Moles = }}\frac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}\]$

Therefore, moles of glucose in the solution can be calculated as:

$\[\begin{gathered} Moles = \frac{{36\,g}}{{180.156\,g/mol}} \hfill \\ Moles = 0.199 \simeq 0.2\,moles \hfill \\ \end{gathered} \]$

Thus, the molality of the solution is given as:

$\[\begin{gathered} Molality = \frac{{0.2}}{{0.250}} \hfill \\ Molality = 0.8 \hfill \\ \end{gathered} \]$

Hence, the molality of the given solution is $0.8$.

(c)

(There are some errors in the question. The correct question will be written as: Molality of 90% by volume $\[{H_2}S{O_4}\]$ (sp. gravity $= 1.98 g/mL$))

Given:

Concentration of solution $\[ = 90\% \]$ by volume

Specific gravity $\[ = 1.98\,g/mL\]$

To Find: Molality of the solution

Solution:

$\[90\% \]$ by volume means in $\[100\,mL\]$ of solution, $\[90\,g\]$ of solute is present.

Therefore,

Mass of solute $\[ = 90\,g\]$

Volume of solution $\[ = 100\,mL\]$

The given solute is $\[{H_2}S{O_4}\]$, therefore, its molar mass is $\[98\,g/mol\]$.

Thus, the number of moles of $\[{H_2}S{O_4}\]$ can be given as:

$\[\begin{gathered} Moles = \frac{{90\,g}}{{98\,g/mol}} \hfill \\ Moles = 0.918 \hfill \\ \end{gathered} \]$

The weight of a solution is given as:

$\[Weight = Volume \times Specific\,gravity\]$

Therefore,

Therefore, the weight of the given solution is

$\[\begin{gathered} Weight = 100\,mL \times 1.98\,g/mL \hfill \\ Weight = 198\,g \hfill \\ \end{gathered} \]$

Thus, the weight of water in the solution can be calculated as:

Weight of water $\[ = 198 - 90 = 108\,g\]$

The molality of a solution can be given as:

$\[{\text{Molality = }}\frac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Weight}}\,{\text{of}}\,{\text{solvent}}\,{\text{(in}}\,{\text{kg)}}}}\]$

Therefore, the molality of the given solution is given as:

$\[\begin{gathered} Molality = \frac{{0.918}}{{108/1000}} \hfill \\ Molality = 8.50 \hfill \\ \end{gathered} \]$

Hence, the molality of the given solution is $8.50$.

(d)

(There are some errors in the question. The correct question will be written as: Calculate the percentage of aluminium in $\[A{l_2}{O_3}\]$ (Atomic wt. of AI = 27, 0 = 16 )​ )

Given:

Aluminium oxide $\[A{l_2}{O_3}\]$

Atomic wt. of $Al = 27$, $O = 16$

To Find: Percentage of aluminium in $\[A{l_2}{O_3}\]$

Solution:

Mass of aluminium in $\[A{l_2}{O_3}\]$ $\[ = 2 \times 17 = 54\]$

Mass of $\[A{l_2}{O_3}\]$ $\[ = 2 \times 27 + 3 \times 16 = 102\,g\]$

The percentage of aluminium in $\[A{l_2}{O_3}\]$ can be given as:

$\[{\text{\% }}\,{\text{of}}\,{\text{Al = }}\frac{{{\text{Mass}}\,{\text{of}}\,{\text{Al}}\,{\text{in}}\,{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{A}}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,}} \times 100\]$

Therefore,

$\[\begin{gathered} \% \,of\,Al = \frac{{54}}{{102}} \times 100 \hfill \\ \% \,of\,Al = 52.94\% \hfill \\ \end{gathered} \]$

Hence, the Percentage of aluminium in $\[A{l_2}{O_3}\]$ is $\[52.94\% \]$.

#SPJ2