Question

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 0.052 m cabr₂ 1.128 m c₂h₅oh 0.30 m nacl 0.073 m mg(no₃)₂

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 - electrolytes, $i = 1$ as they do not dissociate. For ionic compounds, it is equal to the number of ions formed per formula unit upon complete dissociation.

Step2: Analyze $\ce{C2H5OH}$

Ethanol ($\ce{C2H5OH}$) is a non - electrolyte. It does not dissociate into ions in solution. So, for $1.128\ M\ \ce{C2H5OH}$, $i = 1$.

Step3: Analyze $\ce{NaCl}$

Sodium chloride ($\ce{NaCl}$) dissociates as $\ce{NaCl
ightarrow Na^{+}+Cl^{-}}$. One formula unit of $\ce{NaCl}$ gives 2 ions. So, for $0.30\ M\ \ce{NaCl}$, $i = 2$.

Step4: Analyze $\ce{CaBr2}$

Calcium bromide ($\ce{CaBr2}$) dissociates as $\ce{CaBr2
ightarrow Ca^{2 + } + 2Br^{-}}$. One formula unit of $\ce{CaBr2}$ gives 3 ions. So, for $0.052\ M\ \ce{CaBr2}$, $i = 3$.

Step5: Analyze $\ce{Mg(NO3)2}$

Magnesium nitrate ($\ce{Mg(NO3)2}$) dissociates as $\ce{Mg(NO3)2
ightarrow Mg^{2+}+2NO3^{-}}$. One formula unit of $\ce{Mg(NO3)2}$ gives 3 ions. So, for $0.073\ M\ \ce{Mg(NO3)2}$, $i = 3$.

Answer:

$1.128\ M\ \ce{C2H5OH}: i = 1$
$0.30\ M\ \ce{NaCl}: i = 2$
$0.052\ M\ \ce{CaBr2}: i = 3$
$0.073\ M\ \ce{Mg(NO3)2}: i = 3$