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Preparations and Uses of Sodium Carbonate

A compound of...

Question

A compound of sodium does not give $C O_{2}$ when heated but it gives $C O_{2}$ when treated with dilute acids. A crystalline compound is found to have 37.1% Na and 14.52% $H_{2} O$ . Hence, the compound is :

N a H C O_{3} .10 H_{2} O

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N a H C O_{3} .5 H_{2} O

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N a_{2} C O_{3} .10 H_{2} O

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N a_{2} C O_{3} . H_{2} O

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Solution

The correct option is C $N a_{2} C O_{3} . H_{2} O$
$C O_{2}$ is obtained only when the compound is decomposed by acid thus, it is $N a_{2} C O_{3}$ . It cannot be $N a H C O_{3}$ since, it gives $C O_{2}$ on heating. Probable compound is $N a_{2} C O_{3} . x H_{2} O$
percentage Molar ratio Ratio
$N a_{2} C O_{3}$ 85.48 0.8064 1
$H_{2} O$ 14.52 0.8064 1
Thus, compound is $N a_{2} C O_{3} . H_{2} O$

Molecular weight = 124 g mol $^{- 1}$

Percentage of sodium = $\frac{2 \times 23}{124} \times 100 = 37.1 %$

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A compound of sodium does not give CO2 when heated but it gives CO2 when treated with dilute acids. A crystalline compound is found to have 37.1% Na and 14.52% H2O. Hence, the compound is :

A compound of sodium does not give $C O_{2}$ when heated but it gives $C O_{2}$ when treated with dilute acids. A crystalline compound is found to have 37.1% Na and 14.52% $H_{2} O$ . Hence, the compound is :