Q.74. At 298 K, the molar conductivity of x% (w/w) MX solution (aqueous) is 123.5 S cm mol. The cond

Q: What is the correct answer for JEE Main Chemistry Q74?

The correct answer is 15. Given: Molar conductivity ($\Lambda_m$) = 123.5 S cm$^2$ mol$^{-1}$ Conductance (G) = $1.9 \times 10^{-3}$ S Cell constant ($G^*$) = 1.3 cm$^{-1}$ Molar mass of MX = 75 g mol$^{-1}$ Density of solution = 1.0 g mL$^{-1}$ First, calculate the conductivity ($\kappa$) of the solution: $\kappa = G \times G^*$ $\kappa = (1.9 \times 10^{-3} \text{ S}) \times (1.3 \text{ cm}^{-1})$ $\kappa = 2.47 \times

Q: Q.74. At 298 K, the molar conductivity of x% (w/w) MX solution (aqueous) is 123.5 S cm mol. The conductance of same solution is S. The value of x is ______ . (Given : cell constant = 1.3 cm; molar…

The correct answer is 15. Given: Molar conductivity ($\Lambda_m$) = 123.5 S cm$^2$ mol$^{-1}$ Conductance (G) = $1.9 \times 10^{-3}$ S Cell constant ($G^*$) = 1.3 cm$^{-1}$ Molar mass of MX = 75 g mol$^{-1}$ Density of solution = 1.0 g mL$^{-1}$ First, calculate the conductivity ($\kappa$) of the solution: $\kappa = G \times G^*$ $\kappa = (1.9 \times 10^{-3} \text{ S}) \times (1.3 \text{ cm}^{-1})$ $\kappa = 2.47 \times

Question

Q.74. At 298 K, the molar conductivity of x% (w/w) MX solution (aqueous) is 123.5 S cm mol. The conductance of same solution is  S. The value of x is ______ . (Given : cell constant = 1.3 cm; molar mass of MX is 75 g mol, density of aqueous solution of MX at 298 K is 1.0 g mL)

FirstInTest · Accepted Answer

Given:
Molar conductivity ($\Lambda_m$) = 123.5 S cm$^2$ mol$^{-1}$
Conductance (G) = $1.9 	imes 10^{-3}$ S
Cell constant ($G^*$) = 1.3 cm$^{-1}$
Molar mass of MX = 75 g mol$^{-1}$
Density of solution = 1.0 g mL$^{-1}$

First, calculate the conductivity ($\kappa$) of the solution:
$\kappa = G 	imes G^*$
$\kappa = (1.9 	imes 10^{-3} 	ext{ S}) 	imes (1.3 	ext{ cm}^{-1})$
$\kappa = 2.47 	imes 10^{-3} 	ext{ S cm}^{-1}$

Now, relate molar conductivity to conductivity and concentration (C):
$\Lambda_m = \frac{\kappa 	imes 1000}{C}$ (where C is in mol L$^{-1}$ and $\kappa$ in S cm$^{-1}$)
$123.5 = \frac{(2.47 	imes 10^{-3}) 	imes 1000}{C}$
$123.5 = \frac{2.47}{C}$
$C = \frac{2.47}{123.5} = 0.02 	ext{ mol L}^{-1}$

Now, calculate the mass of MX in 1 L of solution:
Mass of MX = $C 	imes 	ext{Molar mass} 	imes 	ext{Volume}$
Mass of MX = $0.02 	ext{ mol L}^{-1} 	imes 75 	ext{ g mol}^{-1} 	imes 1 	ext{ L} = 1.5 	ext{ g}$

Next, calculate the mass of the solution:
Density = 1.0 g mL$^{-1}$
Volume of solution = 1 L = 1000 mL
Mass of solution = Density $	imes$ Volume = $1.0 	ext{ g mL}^{-1} 	imes 1000 	ext{ mL} = 1000 	ext{ g}$

Finally, calculate x% (w/w):
x% (w/w) = $\frac{	ext{mass of solute}}{	ext{mass of solution}} 	imes 100$
x% = $\frac{1.5 	ext{ g}}{1000 	ext{ g}} 	imes 100 = 0.15$

The question asks for the value of x in the form $X 	imes 10^{-2}$.
$0.15 = 15 	imes 10^{-2}$
So, the value of x is 15.

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