determine the vant hoff factor (i) for the following solutions. assume 100% dissociation for ionic…

Explanation:

Step1: Recall van't Hoff factor concept

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

Step2: Analyze $1.128\ M\ C_2H_5OH$

Ethanol ($C_2H_5OH$) is a non - electrolyte. It does not dissociate into ions in solution. So, $i = 1$.

Step3: Analyze $0.30\ M\ NaCl$

Sodium chloride ($NaCl$) dissociates as $NaCl\rightarrow Na^++Cl^-$. One formula unit of $NaCl$ gives 2 ions. So, $i = 2$.

Step4: Analyze $0.0032\ M\ CaBr_2$ and $0.073\ M\ Mg(NO_3)_2$

Calcium bromide ($CaBr_2$) dissociates as $CaBr_2\rightarrow Ca^{2 +}+2Br^-$ and magnesium nitrate ($Mg(NO_3)_2$) dissociates as $Mg(NO_3)_2\rightarrow Mg^{2+}+2NO_3^-$. Both give 3 ions per formula unit. So, $i = 3$.

Answer:

$i = 1$: $1.128\ M\ C_2H_5OH$ $i = 2$: $0.30\ M\ NaCl$ $i = 3$: $0.0032\ M\ CaBr_2$, $0.073\ M\ Mg(NO_3)_2$ $i = 4$: None of the given solutions