A solution consists of ether and 5.0 g of wa | vedclass

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(C) Latent heat of fusion of water $(L_f)$ $= 80 \ cal/g$. Latent heat of vaporization of ether $(L_v)$ $= 90 \ cal/g$. Let the mass of the ether be $

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$4.5$

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(C) Latent heat of fusion of water $(L_f)$ $= 80 \ cal/g$. Latent heat of vaporization of ether $(L_v)$ $= 90 \ cal/g$. Let the mass of the ether be $m$. To freeze $5 \ g$ of water at $0^{\circ} C$, the heat removed from the water is $Q = m_{water} 	imes L_f = 5 \ g 	imes 80 \ cal/g = 400 \ cal$. This heat is absorbed by the ether as it evaporates: $Q = m 	imes L_v$. Equating the heat: $400 \ cal = m 	imes 90 \ cal/g$. $m = \frac{400}{90} \ g \approx 4.44 \ g$. Rounding to the nearest provided option, the mass is approximately $4.5 \ g$.

$A$ solution consists of ether and $5.0 \ g$ of water at $0^{\circ} C$. If the ether evaporates completely to freeze the water, then the mass of the ether in the solution is (in $g$)

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