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After 0.600 L of Ar at 1.29 atm and 223 degrees Celsius is mixed with 0.200 L of O2 at 535 torr and 106 degrees Celsius in a 400.-mL flask at 23 degrees Celsius, what is the pressure in the flask?
Concept of is to be applied here. Let the partial pressure of $Ar$ in the final mixture be $p_"Ar"$ Initially before mixing $ Ar$ had Pressure $P_"Ar"=1.29$ atm Temperature$T_"Ar"=223+273=500$...
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Why does an 0.100 m solution of HCl dissolve in benzene have a freezing point depression of 0.512°C, while an 0.100 m solution of HCl in water has a freezing point depression of 0.372°C?
The reason is that HCl ionizes in water but not in benzene. Freezing point is a colligative property. It depends on the number of particles in solution. The formula for...
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What is the freezing point of a 0.05500 m aqueous solution of $NaNO_3$, if the molal freezing-point-depression constant of water is 1.86°C /m?
We know the depression of freezing point $DeltaT_f=K_fxxbxxi$ Where $K_f->"cryoscopic consrant"=1.86^@"C/m"$ $b->"molal concentration"=0.055m$ $i->"Van'tHoff factor"=2" for "NaNO_3 "*"$ $"*"$ As $NaNO_3$ dissociates as follows producing 2 ions per molecule....
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Consider 1.60 g of naphthalene (C10H8) is dissolved in 20.0 g of
benzene. The freezing point of benzene is 5.5oC, and the freezing point of the
mixture is 2.8oC. What is the molal freezing point constant for benzene?
The formula for calculating freezing point depression is $color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = K_fm color(white)(a/a)|)))" "$ where $ΔT_"f"$ is the decrease in freezing point $K_"f"$ is the molal...
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Calculate the molality of a solution made by dissolving a compound in camphor if the freezing point of the solution is 162.33 degrees C: the melting point of pure camphor was measured to be 178.43 degrees C. K sub f for camphor is 37.7 degrees C/m sub c?
To make any sense of this question, let's first write down what we are given. Given - $color(green)("Pure camphor freezing point" = 178.43^@C)$ - $color(green)(K_(f) = 37.7^@C/m)$ - $color(green)("Solution's freezing...
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An organic solute was dissolved in napthalene such that the freezing point dropped by 7.9 degrees. What is the molality of the solution?
$"1.13 mol/kg"$ to three sig figs (although you only have two...). Well, you either haven't given us the freezing point depression constant, or you weren't given it....
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A solution that contains $"13.2 g"$ of solute in $"250 g"$ of $"CCl"_4$ freezes at $-33^@ "C"$. What is the FW of the solute? $T_f^"*"$ of $"CCl"_4$ is $-22.8^@ "C"$ and its $K_f = -29.8^@ "C/m"$.
$M = "154.26 g/mol"$ You can think of this as just an extension of freezing point depression into solvents that aren't water. Thus, refer to the freezing...
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When $"18 g"$ of ethylene glycol ($C_2H_6O_2$) is dissolved in $"150 g"$ of pure water, what is the freezing point of the solution? (The freezing point depression constant for water is $1.86^@ "C"cdot"kg/mol"$.)
Ethylene glycol is a molecular species, and we need to calculate the of its water solution..... $"Molality"="Moles of solute"/"Kilograms of solvent"=((18*g)/(62.07*g*mol^-1))/((150*g)/(1000*g*kg^-1))=1.93*mol*kg^-1$ $DeltaT_f=k_fxx1.93*mol*kg^-1=(1.86*""^@C*kg)/(mol)xx1.93*(mol)/(kg^-1)$ $DeltaT_f=3.59$ $""^@C$. And this is freezing point depression,...
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Assume that you have a solution of an unknown solute in cyclohexane. If the solution has a freezing-point depression of 9.50Celcius, what is the molality of this solution? (The molal freezing-point constant of cyclohexane is 20.2 C/m)
We obtain a dependent on the assumed van't Hoff factor of $i ~~ 1$, $m ~~ 0.470$ $"molal"$ We refer to the freezing point depression given by...
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Given that the freezing point depression constant for water is 1.86 C kg/mol, how do you calculate the change in freezing point for a 0.833m sugar solution?
$"Freezing point depression"=k_fxx"molality of solution"$ ...where $k_f=underbrace(1.86*K*kg*mol^-1)_"molal freezing point depression constant"$ And so the observed molal freezing point depression is... $1.86*K*kg*mol^-1xx0.833*mol*kg^-1=+1.55*K$.. But this is a freezing point depression, and so...
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