The hydrated salt of $N a_{2} C O_{3} . x H_{2} O$ undergoes $63 %$ loss in mass on heating and become anhydrous. The value of $x$ is:

3

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10

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Solution

The correct option is B $10$
Let the mass of salt is w, then
$w = 106 + 18 x . . . (1)$
after heating $37 w / 100 = 106$
$w = 106 \times 100 / 37$
put value of w in equation (1) we get ;
$106 \times 63 / 37 = 18 x$
$∴ x = 10.02 \approx 10$ .

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

The correct option is B 10Let the mass of salt is w, thenw=106+18x...(1)after heating 37w/100=106w=106×100/37put value of w in equation (1) we get ;106×63/37=18x∴x=10.02≈10.

The hydrated salt of $N a_{2} C O_{3} . x H_{2} O$ undergoes $63 %$ loss in mass on heating and become anhydrous. The value of $x$ is:

The correct option is B $10$
Let the mass of salt is w, then
$w = 106 + 18 x . . . (1)$
after heating $37 w / 100 = 106$
$w = 106 \times 100 / 37$
put value of w in equation (1) we get ;
$106 \times 63 / 37 = 18 x$
$∴ x = 10.02 \approx 10$ .