5 milli-moles of a solid A was dissolved in 5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The rate of polymerization is zero order: $[A]_t = [A]_0 - Kt$.Given $[A]_0 = 5 	imes 10^{-3} \, moles$,$t = 20 \, minutes$,and $K = 10^{-X} \, m

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The rate of polymerization is zero order: $[A]_t = [A]_0 - Kt$.Given $[A]_0 = 5 	imes 10^{-3} \, moles$,$t = 20 \, minutes$,and $K = 10^{-X} \, moles/minute$.Remaining moles of $A$ after $20$ minutes: $n_A = 5 	imes 10^{-3} - 20 	imes 10^{-X}$.After adding $6 	imes 10^{-3} \, moles$ of $C$,the total moles of solute in $5 \, moles$ $(0.005 \, kg)$ of water is $n_{total} = (5 	imes 10^{-3} - 20 	imes 10^{-X}) + 6 	imes 10^{-3} = 11 	imes 10^{-3} - 20 	imes 10^{-X}$.Using the depression of freezing point formula: $\Delta T_f = K_f \cdot m$.$0.186 = 1.86 \cdot \frac{n_{total}}{0.005 \, kg}$ $\Rightarrow 0.1 = \frac{n_{total}}{0.005}$ $\Rightarrow n_{total} = 0.0005 \, moles$.Wait,recalculating based on molality $m = \frac{n_{total}}{0.005} = 0.1 \Rightarrow n_{total} = 0.1 	imes 0.005 = 0.0005 \, moles$.Setting $11 	imes 10^{-3} - 20 	imes 10^{-X} = 0.5 	imes 10^{-3}$ $\Rightarrow 20 	imes 10^{-X} = 10.5 	imes 10^{-3} = 1.05 	imes 10^{-2}$.$10^{-X} = 0.0525 	imes 10^{-2} = 5.25 	imes 10^{-4}$.Re-evaluating the provided solution logic: $0.1 = \frac{n_{total}}{0.090 \, kg}$ (assuming $5 \, moles$ of water is $0.090 \, kg$ is incorrect,$5 \, moles = 0.090 \, kg$ is $90 \, g$).Given the provided solution steps: $0.1 = \frac{11 	imes 10^{-3} - 20 	imes 10^{-X}}{0.090}$ $\Rightarrow 0.009 = 0.011 - 20 	imes 10^{-X}$ $\Rightarrow 20 	imes 10^{-X} = 0.002$ $\Rightarrow 10^{-X} = 10^{-4}$ $\Rightarrow X = 4$.

Similar Questions

Derive an expression to calculate the time required for the completion of a zero order reaction.

The decomposition of $NH_3$ on $Pt$ surface is a zero order reaction. If the value of rate constant is $2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$,the rates of appearance of $N_2$ and $H_2$ are respectively:

For a zero-order reaction,the correct expression for rate constant $(k)$ at half-life time $(t_{1/2})$ is ($[R_0] =$ initial concentration of reactant).

Which of the following relations is correct for a zero order reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module