$10 g$ of $N a_{2} C O_{3} \cdot 10 H_{2} O$ is strongly to remove all the water. The percentage reduction in the mass of $N a_{2} C O_{3} \cdot 10 H_{2} O$ will be :

10 %

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!

62.9 %

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

69.2 %

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

Open in App

Solution

The correct option is C $62.9 %$
Molecular mass of $N a_{2} C O_{3} .10 H_{2} O$ is $286 u$ .

Molecular mass of anhydrate $N a_{2} C O_{3}$ is $106 u$ .

Difference in mass $286 u - 106 u = 180 u$

Percentage decrease $= \frac{D i f f e r n c e}{O r i g i n a l} \times 100$

$\Rightarrow % d e c r e a s e = \frac{180}{286} \times 100 = 62.9 %$
Therefore option C $62.9 %$ Is correct.

Suggest Corrections

Similar questions

L i_{2} C O_{3}

N a_{2} C O_{3} \cdot 10 H_{2} O

L i_{2} C O_{3}

N a_{2} C O_{3} \cdot 10 H_{2} O

The formula of soda ash is
(a) $N a_{2} C O_{3} .10 H_{2} O$ (b) $N a_{2} C O_{3} . 2 H_{2} O$
(c) $N a_{2} C O_{3} . H_{2} O$ (d) $N a_{2} C O_{3}$

0.2 m o l e

N a_{2} C O_{3} \cdot 10 H_{2} O

Join BYJU'S Learning Program