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

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(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: $0.1 = \frac{11 	imes 10^{-3} - 20 	imes 10^{-X}}{0.090}$ $\Rightarrow 0.009 = 0.011 - 20 	imes 10^{-X}$ $\Rightarrow 20 	imes 10^{-X} = 0.002$ $\Rightarrow 10^{-X} = 10^{-4}$ $\Rightarrow X = 4$.

$[R] \text{ (molar)}$	$1.0$	$0.75$	$0.40$	$0.10$
$t \text{ (min.)}$	$0.0$	$0.05$	$0.12$	$0.18$

$Time \ (s)$	$Rate \ (mol \ L^{-1} s^{-1})$
$0$	$1.60 \times 10^{-2}$
$10$	$1.60 \times 10^{-2}$
$20$	$1.60 \times 10^{-2}$
$30$	$1.60 \times 10^{-2}$

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