Question

question 5 of 29
determine the vant hoff factor (i) for the following solutions. assume 100% dissociation for ionic solutes.
i = 1
i = 2
i = 3
i = 4
answer bank
1.128 m c₂h₅oh 0.073 m mg(no₃)₂ 0.30 m nacl 0.0032 m cabr₂

Explanation:

Step1: Identify non - ionic solute

Ethanol ($C_2H_5OH$) is a non - ionic compound. Non - ionic compounds do not dissociate in solution, so the van't Hoff factor $i = 1$.

Step2: Analyze $Mg(NO_3)_2$ dissociation

$Mg(NO_3)_2$ dissociates as $Mg(NO_3)_2
ightarrow Mg^{2 + }+2NO_3^{-}$. One formula unit gives 3 ions, so $i = 3$.

Step3: Analyze $NaCl$ dissociation

$NaCl$ dissociates as $NaCl
ightarrow Na^{+}+Cl^{-}$. One formula unit gives 2 ions, so $i = 2$.

Step4: Analyze $CaBr_2$ dissociation

$CaBr_2$ dissociates as $CaBr_2
ightarrow Ca^{2 + }+2Br^{-}$. One formula unit gives 3 ions, so $i = 3$.

Answer:

1.128 M $C_2H_5OH$: $i = 1$
0.073 M $Mg(NO_3)_2$: $i = 3$
0.30 M $NaCl$: $i = 2$
0.0032 M $CaBr_2$: $i = 3$