Step 1: Define the van't Hoff factor. The van't Hoff factor (\( i \)) quantifies the number of particles a compound yields upon dissolution. For ionic substances, \( i \) is contingent upon the extent of dissociation. Step 2: Evaluate the provided compounds. 1. KCl dissociates into \( K^+ \) and \( Cl^- \), resulting in 2 particles, thus \( i = 2 \). 2. NaCl dissociates into \( Na^+ \) and \( Cl^- \), also yielding 2 particles, so \( i = 2 \). 3. \( K_2SO_4 \) dissociates into \( 2K^+ \) and \( SO_4^{2-} \), producing a total of 3 particles, hence \( i = 3 \). Step 3: Conclude the findings. The van't Hoff factors for KCl, NaCl, and \( K_2SO_4 \) are 2, 2, and 3, respectively.
If the molar conductivity ($\Lambda_m$) of a 0.050 mol $L^{–1}$ solution of a monobasic weak acid is 90 S $cm^{2} mol^{–1}$, its extent (degree) of dissociation will be:
[Assume: $\Lambda^0$ = 349.6 S $cm^{2} mol^{–1}$ and $\Lambda^0_{\text{acid}}$ = 50.4 S$ cm^{2} mol^{–1}$]