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Raoult's Law

1 mole of a m...

Question

1 mole of a mixture of $L i_{2} C O_{3}$ and $N a_{2} C O_{3}$ forms 0.75 mole of $C O_{2}$ at NTP. What is percentage of each in the mixture ?

A

37 m o l % L i_{2} C O_{3}

63 m o l % N a_{2} C O_{3}

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B

63 m o l % L i_{2} C O_{3}

37 m o l % N a_{2} C O_{3}

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C

75 m o l % L i_{2} C O_{3}

25 m o l % N a_{2} C O_{3}

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D

25 m o l % L i_{2} C O_{3}

75 m o l % N a_{2} C O_{3}

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Solution

The correct option is C $75 m o l % L i_{2} C O_{3}$ $25 m o l % N a_{2} C O_{3}$
$L i_{2} C O_{3} Δ \to L i_{2} O + C O_{2}$
$N a_{2} C O_{3} Δ \to X (n o e f f e c t)$
$C O_{2}$ is due to $L i_{2} C O_{3}$ only
hence, 0.75 mol $C O_{2} = 0.75 m o l$ of $L i_{2} C O_{3}$ $75 m o l % L i_{2} C O_{3}$
$25 m o l % N a_{2} C O_{3}$

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