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:
(Gram atomic mass : $N a = 23; C = 12; O = 16$ )

10

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $10$
$M_{N a_{2} C O_{3}} = 106$
$M_{H_{2} O} = 18$

Let the mass of salt is $W$ .
Then $106 + 18 n = W ⟶ (1)$

After heating, $\frac{37 W}{100} = 106 ⟶ (2)$ ( $63$ % is lost i.e, $H_{2} O$ is lost)

Putting (2) in (1),
$\Rightarrow 106 + 18 n = \frac{106 \times 100}{37}$
$\Rightarrow 18 n = 180.4$

$\Rightarrow n = 10.02 ≃ 10$

Hence, option $A$ is correct.

Suggest Corrections

Similar questions

The hydrated salt Na2CO3.xH2O undergoes 63% loss in mass on heating and becomes anhydrous .The value of x is (pls give explanation for necessary steps)

Q. 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: (report the answer to the closest integer)

Q. 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:

Q. The hydrated salt

N a_{2} S O_{4}, 10 H_{2} O

undergoes

X %

loss in weight on heating and becomes anhydrous. The value of

X

will be:

Q. The hydrated salt,

N a_{2} S O_{4} . n H_{2} O

undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of

n

will be:

Join BYJU'S Learning Program

The hydrated salt Na2CO3.xH2O undergoes 63% loss in mass on heating and becomes anhydrous. The value of x is: (Gram atomic mass : Na=23;C=12;O=16)

The correct option is A 10MNa2CO3=106MH2O=18Let the mass of salt is W. Then 106+18n=W⟶(1) After heating, 37W100=106⟶(2) (63% is lost i.e, H2O is lost)Putting (2) in (1),⇒106+18n=106×10037⇒18n=180.4⇒n=10.02≃10Hence, option A is correct.

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:
(Gram atomic mass : $N a = 23; C = 12; O = 16$ )