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# Hi! I cannot figure these 2 out for the life of me. For the boiling point, I thought the equation was dTb=(i)(kb)(m) but perhaps I am using it wrong. Any help would be appreciated.$52.9 \mathrm{~g}$ of $\mathrm{CuCl}_{2}(\mathrm{MW}=134.45)$ is dissolved in $800.0 \mathrm{~mL}$ of water at $52^{\circ} \mathrm{C}$. The solution is maintained at a temperature of $52^{\circ} \mathrm{C}$. The vapor pressure of pure water at $52^{\circ} \mathrm{C}$ is 102.1 torr and the density of water $(M W=18.05)$ at $52^{\circ} \mathrm{C}$ is $0.987 \mathrm{~g} / \mathrm{mL}$.4 $0 / 2$ points What is the boiling point of the solution is $k_{b}$ for water is 0.51 (K $\mathrm{kg}$ )/mol? (Round to two decimal places.) $\times 100.75{ }^{\circ} \mathrm{C}$ 5 0/2 points What mass of $\mathrm{NaCl}_{(\mathrm{s})}$ (MW 58.44) would have to be added to $800 \mathrm{~mL}$ of water so that the resulting solution has the same freezing point as the $\mathrm{CuCl}_{2(\mathrm{aq})}$ solution described above? Assume ideal solutions. (Round to one decimal place.)

## Public Answer

WMHBZF The First Answerer