Question

How many kilograms of wet ${\text{NaOH}}$ containing $10\% $ water are required to prepare $60{\text{ litre}}$ of a $0.5{\text{ M}}$ solution?
A. $1.23{\text{ kg}}$
B. $1.13{\text{ kg}}$
C. $1.33{\text{ kg}}$
D. $1.24{\text{ kg}}$

Answer 1

Hint: Follow the given steps to calculate the kilograms of wet ${\text{NaOH}}$ containing $10\% $ water required to prepare $60{\text{ litre}}$ of a $0.5{\text{ M}}$ solution.
1Calculate the number of moles of ${\text{NaOH}}$ in $60{\text{ litre}}$ of solution.
2Calculate the amount of ${\text{NaOH}}$ in the calculated number of moles of ${\text{NaOH}}$ using the molar mass.
3As the water contained in ${\text{NaOH}}$ is $10\% $, calculate the kilograms of pure ${\text{NaOH}}$ in it.

Complete step by step answer:
Calculate the number of moles of ${\text{NaOH}}$ in $60{\text{ litre}}$ of solution as follows:
The molarity of the ${\text{NaOH}}$ solution is $0.5{\text{ M}}$. The molarity of a solution is the number of moles of solute in $1{\text{ litre}}$ of solution. Thus, the molarity of the ${\text{NaOH}}$ solution is $0.5{\text{ mol liter}}{{\text{e}}^{ - 1}}$.
Calculate the number of moles ${\text{NaOH}}$ in $60{\text{ litre}}$ of solution using the relation as follows:
${\text{Number of moles}} = {\text{Molarity}} \times {\text{Volume of solution}}$
Substitute $0.5{\text{ mol liter}}{{\text{e}}^{ - 1}}$ for the molarity of the ${\text{NaOH}}$ solution, $60{\text{ litre}}$ for the volume of the solution. Thus,
${\text{Number of moles of NaOH}} = 0.5{\text{ mol }}{{{\text{litr}}{{\text{e}}^{ - 1}}}} \times 60{\text{ }}{{{\text{litre}}}}$
${\text{Number of moles of NaOH}} = 30{\text{ mol}}$
Thus, the number of moles of ${\text{NaOH}}$ in $60{\text{ litre}}$ of solution are $30{\text{ mol}}$.
Step 2:
Calculate the molar mass of ${\text{NaOH}}$ as follows:
\[{\text{Molar mass of NaOH}} = \left( {{\text{1}} \times {\text{Atomic mass of Na}}} \right) + \left( {{\text{1}} \times {\text{Atomic mass of O}}} \right) + \left( {{\text{1}} \times {\text{Atomic mass of H}}} \right)\]
\[{\text{Molar mass of NaOH}} = \left( {1 \times 23} \right) + \left( {1 \times 16} \right) + \left( {1 \times 1} \right)\]
\[{\text{Molar mass of NaOH}} = 23 + 16 + 1\]
\[{\text{Molar mass of NaOH}} = 40{\text{ g mo}}{{\text{l}}^{ - 1}}\]
Thus, the molar mass of ${\text{NaOH}}$ is \[40{\text{ g mo}}{{\text{l}}^{ - 1}}\].
Step 3:
Calculate the amount of ${\text{NaOH}}$ in $30{\text{ mol}}$ of ${\text{NaOH}}$ as follows:
$1{\text{ mol}}$ of a substance contains an amount of a substance in grams equal to its molar mass.
Thus, $1{\text{ mol}}$ of ${\text{NaOH}}$ contains \[40{\text{ g}}\] of ${\text{NaOH}}$.
Thus, $30{\text{ mol}}$ of ${\text{NaOH}}$ contains,
${\text{Amount of NaOH}} = 30{\text{ }}{{{\text{mol NaOH}}}} \times \dfrac{{40{\text{ g NaOH}}}}{{{\text{1 }}{{{\text{mol NaOH}}}}}}$
${\text{Amount of NaOH}} = 1200{\text{ g NaOH}} = 1200 \times {10^{ - 3}}{\text{ kg NaOH}}$
Thus, the amount of ${\text{NaOH}}$ in $30{\text{ mol}}$ of ${\text{NaOH}}$ is $1200 \times {10^{ - 3}}{\text{ kg}}$.
Step 4:
Calculate the kilograms of ${\text{NaOH}}$ containing $10\% $ water required to prepare $60{\text{ litre}}$ of a $0.5{\text{ M}}$ solution as follows:
The ${\text{NaOH}}$ contains $10\% $ water. Thus, in $100{\text{ kg}}$ ${\text{NaOH}}$, pure ${\text{NaOH}}$ is $\left( {100 - 10} \right) = 90{\text{ kg}}$.
$100{\text{ kg}}$ ${\text{NaOH}}$ contains $90{\text{ kg}}$ pure ${\text{NaOH}}$
Thus, $1200 \times {10^{ - 3}}{\text{ kg}}$ ${\text{NaOH}}$ contains,
${\text{NaOH }}\left( {{\text{kg}}} \right) = \dfrac{{100{\text{ }}{{{\text{kg}}}} \times 1200 \times {{10}^{ - 3}}{\text{ kg}}}}{{90{\text{ }}{{{\text{kg}}}}}}$
${\text{NaOH }}\left( {{\text{kg}}} \right) = 1.33{\text{ kg}}$
Thus, the kilograms of ${\text{NaOH}}$ containing $10\% $ water required to prepare $60{\text{ litre}}$ of a $0.5{\text{ M}}$ solution are $1.33{\text{ kg}}$.

So, the correct answer is OptionC.

Note:
The ${\text{NaOH}}$ contains $10\% $ water. Thus, the pure ${\text{NaOH}}$ is $90\% $. Thus, calculate the kilograms of pure ${\text{NaOH}}$.