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(A) The boiling point of water at $1 \ atm$ is $100 \ ^oC$. The elevation in boiling point is $\Delta T_b = 110 \ ^oC - 100 \ ^oC = 10 \ K$.$CaBr_2$ d

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$1400$

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(A) The boiling point of water at $1 \ atm$ is $100 \ ^oC$. The elevation in boiling point is $\Delta T_b = 110 \ ^oC - 100 \ ^oC = 10 \ K$.$CaBr_2$ dissociates as $CaBr_2 ightarrow Ca^{2+} + 2Br^-$,so the van't Hoff factor $i = 3$.The initial mass of the solution is $2.8 \ kg = 2800 \ g$. The molality $m = 2 \ mol/kg$. Let $W_s$ be the mass of solute and $W_w$ be the mass of solvent (water).$m = \frac{n_{solute}}{W_w (in \ kg)}$ $\Rightarrow 2 = \frac{n_{solute}}{W_w}$ $\Rightarrow n_{solute} = 2 W_w$.Total mass $= W_s + W_w = 2800 \ g$. Molar mass of $CaBr_2 = 40 + 2(80) = 200 \ g/mol$.$W_s = n_{solute} 	imes 200 = (2 W_w) 	imes 200 = 400 W_w$.$400 W_w + W_w = 2800$ $\Rightarrow 401 W_w = 2800$ $\Rightarrow W_w \approx 6.98 \ kg$ (This implies the initial solution was not $2.8 \ kg$ for $2 \ m$ concentration). Re-evaluating based on the standard formula $\Delta T_b = K_b 	imes m_{final} 	imes i$:$10 = 0.5 	imes m_{final} 	imes 3 \Rightarrow m_{final} = \frac{10}{1.5} = 6.67 \ mol/kg$.Initial moles of $CaBr_2 = 2 \ mol/kg 	imes 2 \ kg = 4 \ mol$ (assuming $2 \ kg$ solvent). Mass of $CaBr_2 = 4 	imes 200 = 800 \ g$. Initial water $= 2000 \ g$.$6.67 = \frac{4}{W_{final}} \Rightarrow W_{final} = 0.6 \ kg = 600 \ g$.Water evaporated $= 2000 \ g - 600 \ g = 1400 \ g$.

Calculate the amount of water evaporated if $2.8 \ kg$ of $2 \ m$ $CaBr_2$ solution is heated to $110 \ ^oC$ at $1 \ atm$ .......... $gm$.
(Given: $K_b$ of water $= 0.5 \ K \ kg/mol$; Atomic mass of $Ca = 40$,$Br = 80$)

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Calculate the amount of water evaporated if $2.8 \ kg$ of $2 \ m$ $CaBr_2$ solution is heated to $110 \ ^oC$ at $1 \ atm$ .......... $gm$.(Given: $K_b$ of water $= 0.5 \ K \ kg/mol$; Atomic mass of $Ca = 40$,$Br = 80$)