
The hydrated salt $N{{a}_{2}}S{{O}_{4}}.n{{H}_{2}}O$, undergoes 56% loss in weight on heating and becomes anhydrous. The value of n will be:
A. $5$
B. $3$
C. $7$
D. $10$
Answer
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Hint: To solve such questions, first find out the total weight of the hydrated salt. Remember on heating, the loss in weight is due to the elimination of water molecules from the hydrated salt. So, find out how much will be the total weight after loss and then you can get the n value.
Complete answer:
Given that,
The hydrated salt $N{{a}_{2}}S{{O}_{4}}.n{{H}_{2}}O$, undergoes 56% loss in weight on heating and becomes anhydrous. And, we have to find out the n-value which represents the number of moles of water molecules that are present in the hydrated salt.
On heating the hydrated salt, equal number of moles of water will be formed as the product and the balanced reaction of this procedure can be shown as followed:
$N{{a}_{2}}S{{O}_{4}}.n{{H}_{2}}O\xrightarrow{\Delta }N{{a}_{2}}S{{O}_{4}}+n{{H}_{2}}O$
So, let us find out the molecular weights of the compounds in the reaction to know the n-value.
The atomic weight of sodium is $23g$.
The atomic weight of a sulphur atom is$32g$.
The atomic weight of an oxygen atom is $16g$.
The atomic weight of hydrogen is $1g$.
So, the total molecular weight of $N{{a}_{2}}S{{O}_{4}}.n{{H}_{2}}O$ will be
$2\times 23+32\times 1+4\times 16+n(2\times 1+16)=142+18n$.
Given that, 56% loss in total weight of the salt occurs on heating. We should know that loss in water is mainly responsible for the loss in weight of the salt. So, the percentage loss can be shown as:
%loss= $\dfrac{\text{weight of water}}{\text{Total weight}}\times 100$
And we know percentage loss is already given.
Weight of water is $18n$ and the total weight is $142+18n$.
So, %loss=$\dfrac{18n}{142+18n}\times 100$
Now placing the value of percentage loss, we get:
$56=\dfrac{18n}{142+18n}\times 100$
Then, $1800n=56\times 142+(18\times 56)n$
So, $1800n=7952+1008n$
Then, $792n=7952$
And, $n=10.04\cong 10$
Therefore, the value of n will be $10$.
Hence, the correct option is D.
Note:
The compound with the chemical formula , is referred to as the sodium sulphate. This compound is also usually used to dry various organic liquids. It is also used for the manufacture of detergents and for defrosting windows, in carpet fresheners, starch manufacture, etc.
Complete answer:
Given that,
The hydrated salt $N{{a}_{2}}S{{O}_{4}}.n{{H}_{2}}O$, undergoes 56% loss in weight on heating and becomes anhydrous. And, we have to find out the n-value which represents the number of moles of water molecules that are present in the hydrated salt.
On heating the hydrated salt, equal number of moles of water will be formed as the product and the balanced reaction of this procedure can be shown as followed:
$N{{a}_{2}}S{{O}_{4}}.n{{H}_{2}}O\xrightarrow{\Delta }N{{a}_{2}}S{{O}_{4}}+n{{H}_{2}}O$
So, let us find out the molecular weights of the compounds in the reaction to know the n-value.
The atomic weight of sodium is $23g$.
The atomic weight of a sulphur atom is$32g$.
The atomic weight of an oxygen atom is $16g$.
The atomic weight of hydrogen is $1g$.
So, the total molecular weight of $N{{a}_{2}}S{{O}_{4}}.n{{H}_{2}}O$ will be
$2\times 23+32\times 1+4\times 16+n(2\times 1+16)=142+18n$.
Given that, 56% loss in total weight of the salt occurs on heating. We should know that loss in water is mainly responsible for the loss in weight of the salt. So, the percentage loss can be shown as:
%loss= $\dfrac{\text{weight of water}}{\text{Total weight}}\times 100$
And we know percentage loss is already given.
Weight of water is $18n$ and the total weight is $142+18n$.
So, %loss=$\dfrac{18n}{142+18n}\times 100$
Now placing the value of percentage loss, we get:
$56=\dfrac{18n}{142+18n}\times 100$
Then, $1800n=56\times 142+(18\times 56)n$
So, $1800n=7952+1008n$
Then, $792n=7952$
And, $n=10.04\cong 10$
Therefore, the value of n will be $10$.
Hence, the correct option is D.
Note:
The compound with the chemical formula , is referred to as the sodium sulphate. This compound is also usually used to dry various organic liquids. It is also used for the manufacture of detergents and for defrosting windows, in carpet fresheners, starch manufacture, etc.
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