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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$Na_2[CuF_4]$

Vedclass · Answer

(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:

Similar Questions

The freezing points of equimolar solutions of glucose,$KNO_3$ and $AlCl_3$ are in the order of:

The correct relationship between the boiling points of very dilute solutions of $AlCl_3$ $(t_1)$ and $CaCl_2$ $(t_2)$,having the same molar concentration is

$A$ $0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20\%$ ionized. If $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$,the depression in freezing point of the solution is ....... $K$.

The molal elevation constant of water is $0.51 \ K \, kg \, mol^{-1}$. The boiling point of $0.1 \ m$ aqueous $NaCl$ solution is nearly .......... $^oC$.

Which of the following aqueous solutions has the highest freezing point?

Vedclass Test Series

Exam Paper Generator

Online Exam Module