Question

A 6.85g sample of the hydrated $Sr{(OH)_2}.x{H_2}O$ is dried in an oven to give 3.13g of anhydrous $Sr{(OH)_2}$. What is the value of $x$ ? (Atomic masses: Sr $ = $ 87.60, O $ = $ 16, H $ = $ 1.0)
A.8
B.12
C.10
D.6

Answer 1

Hint: From given question
$Sr{(OH)_2}.x{H_2}O \to Sr{(OH)_2} + x{H_2}O$
Then find out the percentage of ${H_2}O$ by using weight given in the equation and calculate percentage of ${H_2}O$ using molecular weight in $Sr{(OH)_2}$ and compare both percentages. Both percentages must be the same.

Complete step by step answer:
According to the question, let us write the given values.
Given:
Weight of $Sr{(OH)_2}.x{H_2}O$ before drying $ = $ 6.85 g
Weight of anhydrous $Sr{(OH)_2}$ $ = $ 3.13 g
Now, we can write the reaction as follows:
$Sr{(OH)_2}.x{H_2}O \to Sr{(OH)_2} + x{H_2}O$
$6.85g \to 3.13 + ?$
Therefore, weight of \[x{H_2}O\] can be written as
Weight of \[x{H_2}O\] $ = $ Weight of $Sr{(OH)_2}.x{H_2}O$ $ - $ Weight of $Sr{(OH)_2}$
Weight of \[x{H_2}O\] $ = $ 6.85 g $ - $ 3.13 g
Weight of \[x{H_2}O\] $ = $ 3.72 g
So, the weight of water molecules is 3.72 g
Now, by using the weight we can calculate the required percentage
Percentage of \[{H_2}O\] $ = $ $\dfrac{{3.72}}{{6.85}} \times 100 = 54.30\% $
Percentage of \[{H_2}O\] in $Sr{(OH)_2}.x{H_2}O$ $ = \dfrac{{18x}}{{121 + 18x}} \times 100 = 54.30\% $
Now find out the value of $x$ by putting different value given in the options,
First we have $x$ $ = $ 8
Percentage of \[{H_2}O\] $ = \dfrac{{18 \times 8}}{{121 + 18 \times 8}} \times 100 = 54.30\% $
Second we have $x$ $ = $ 12
Percentage of \[{H_2}O\] \[ = \dfrac{{18 \times 12}}{{121 + 18 \times 12}} \times 100 = 64.09\% \]
Third we have $x$ $ = $ 10
Percentage of \[{H_2}O\] \[ = \dfrac{{18 \times 10}}{{121 + 18 \times 10}} \times 100 = 59.80\% \]
Fourth we have $x$ $ = $ 6
Percentage of \[{H_2}O\] \[ = \dfrac{{18 \times 6}}{{121 + 18 \times 6}} \times 100 = 47.16\% \]
From the above calculations we can say that the value of $x$ is 8.

Therefore, the correct option is A.

Note: $Sr{(OH)_2}$ known as strontium hydroxide is more soluble in cold water than in hard water. Due to its solubility in cold water its preparation is easily carried out by addition of strong base such as $NaOH$ or \[KOH\] drop by drop to a solution of any soluble strontium salt.
Strontium hydroxide is used in refining of beet sugar and stabilizer in plastic.
It is used in providing finishing touch for clay portraits.
They have their application in electrical use also.