R=ρAℓκ=G⋅G∗G=R1;κ=ρ1 G∗=RρAℓAℓ= Resistance = Resistivity = cell constant (G∗) κdκc=RcRd;λm=Cκ×1000κdκc=(λm⋅C)d(λm⋅C)=RcRdc= concentrated sol. d = diluted solution 150.(0.1)2100⋅(0.15)2=100RdRd=150Ω
JEE Main 2025 — Chemistry Physical Chemistry
Given below is the plot of the molar conductivity vs concentration for KCl in aqueous solution. 
If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100Ω, then the resistance of the same cell with the dilute solution is ' x ' Ω The value of x is __________ (Nearest integer)
Held on 28 Jan 2025 · Verified 6 Jul 2026.
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80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
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