A compound contains 1.08 mol of Na, 0.539 m | vedclass

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(C) $1$. Determine the empirical formula: The molar ratios are $Na : Cu : F = 1.08 : 0.539 : 2.16$. Dividing by $0.539$,we get $Na : Cu : F \approx 2

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$Na_2[CuF_4]$

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(C) $1$. Determine the empirical formula: The molar ratios are $Na : Cu : F = 1.08 : 0.539 : 2.16$. Dividing by $0.539$,we get $Na : Cu : F \approx 2 : 1 : 4$. Thus,the empirical formula is $Na_2CuF_4$.$2$. Determine the van't Hoff factor $(i)$: The osmotic pressure is three times that of urea (a non-electrolyte,$i=1$),so $i = 3$.$3$. Analyze the dissociation: $A$ compound with $i=3$ dissociates into $3$ ions. $Na_2[CuF_4]$ dissociates as $Na_2[CuF_4] \rightarrow 2Na^+ + [CuF_4]^{2-}$. This gives $2 + 1 = 3$ ions.$4$. Conclusion: The formula is $Na_2[CuF_4]$.

$A$ compound contains $1.08 \ mol$ of $Na, 0.539 \ mol$ of $Cu$ and $2.16 \ mol$ of $F$. Its aqueous solution shows osmotic pressure which is three times that of urea having the same molar concentration. The formula of the compound is:

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