The hydrated salt ${N a}_{2} {C O}_{3} . {x H}_{2} O$ undergoes 63% loss in mass on heating and becomes anhydrous. The value of $x$ is:

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Solution

The correct option is A 10
Let mass of the salt is $w$ , then,
$w = 106 + 18 n$ -----------------1
After heating,
$\frac{37 w}{100} = 106$
$w = \frac{106 \times 100}{37}$
$∴$ From Equation 1 it implies,
$\Rightarrow \frac{106 \times 100}{37} = 106 + 18 n$
$180.48 = 18 n$
$∴ n = 10.02$
So, value of $x = 10$

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The hydrated salt Na2CO3.xH2O undergoes 63% loss in mass on heating and becomes anhydrous. The value of x is:

The correct option is A 10Let mass of the salt is w, then,w=106+18n-----------------1After heating,37w100=106w=106×10037∴ From Equation 1 it implies, ⇒106×10037=106+18n180.48=18n∴n=10.02So, value of x=10

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