Question

question 2 of 29
determine the vant hoff factor (i) for the following solutions. assume 100% dissociation for ionic solutes.
0.35 m nacl; i=
0.0038 m mgbr₂; i=
1.269 m c₆h₁₂o₆; i=
0.087 m ca(clo₄)₂; i=

Explanation:

Step1: Recall van't Hoff factor concept

The van't Hoff factor ($i$) is the number of particles into which a solute dissociates per formula unit in solution. For non - ionic compounds, $i = 1$ as they do not dissociate into ions. For ionic compounds, it is the sum of the ions produced upon dissociation.

Step2: Analyze NaCl

NaCl dissociates as $NaCl
ightarrow Na^{+}+Cl^{-}$. The number of ions produced is 2. So, for 0.35 M NaCl, $i = 2$.

Step3: Analyze $MgBr_{2}$

$MgBr_{2}$ dissociates as $MgBr_{2}
ightarrow Mg^{2 +}+2Br^{-}$. The sum of the ions is $1 + 2=3$. So, for 0.0038 M $MgBr_{2}$, $i = 3$.

Step4: Analyze $C_{6}H_{12}O_{6}$

$C_{6}H_{12}O_{6}$ is a non - ionic compound (a sugar). It does not dissociate into ions in solution. So, for 1.269 M $C_{6}H_{12}O_{6}$, $i = 1$.

Step5: Analyze $Ca(ClO_{4})_{2}$

$Ca(ClO_{4})_{2}$ dissociates as $Ca(ClO_{4})_{2}
ightarrow Ca^{2+}+2ClO_{4}^{-}$. The sum of the ions is $1 + 2 = 3$. So, for 0.087 M $Ca(ClO_{4})_{2}$, $i = 3$.

Answer:

0.35 M NaCl: $i = 2$
0.0038 M $MgBr_{2}$: $i = 3$
1.269 M $C_{6}H_{12}O_{6}$: $i = 1$
0.087 M $Ca(ClO_{4})_{2}$: $i = 3$