Answer
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Hint: We can use Henry’s law to formulate the formula required for the given question which is given as follows
\[\Delta {{T}_{f}}={{K}_{f}}m\]
Where, \[\Delta {{T}_{f}}\] is the change in the freezing point, \[{{K}_{f}}\] is the freezing constant and \[m\]is the molality of the solution in consideration.
Step by step answer:
Let W g of compound is to be dissolved in the benzene.
Number of moles of compound will be given by the following formula
$n=\dfrac{mass}{\text{molar mass}}$
\[n=\dfrac{W}{256}\]
Mass of benzene will be = $75 g$ \[=\dfrac{75}{1000}kg\] = $0.075 kg$
Now we can calculate molality of the solution, $m=\dfrac{moles}{\text{mass of solvent}}$
\[m=\dfrac{W}{256\times 0.075}\]= $19.2W molal$
The depression in the freezing point, \[\Delta {{T}_{f}}\] = $0.48K$
The depression in the freezing point constant = \[{{K}_{f}}\]= $5.12 Kg/mol$
\[\Delta {{T}_{f}}={{K}_{f}}m\]
Now, substituting these values in the given formula we get,
\[0.48=5.12\times \dfrac{W}{19.6}\]
\[W=\dfrac{0.48\times 19.2}{5.12}=1.8\]
Hence the correct answer is that $1.8 g$ of solute is to be dissolved.
Additional information:
Henry's law is defined as the gas laws formulated by William Henry in 1803 and it states that at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid. We can write it mathematically by stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid:
$C$ = $k[P_{gas}]$
Where, $C$ is the solubility of a gas at a fixed temperature in a particular solvent, $k$ is Henry's law constant and $[P_{gas}]$ is the partial pressure of the gas
Note: Freezing point can be defined as the temperature at which a liquid changes to a solid. The freezing point is lower than the melting point in the case of mixtures.
\[\Delta {{T}_{f}}={{K}_{f}}m\]
Where, \[\Delta {{T}_{f}}\] is the change in the freezing point, \[{{K}_{f}}\] is the freezing constant and \[m\]is the molality of the solution in consideration.
Step by step answer:
Let W g of compound is to be dissolved in the benzene.
Number of moles of compound will be given by the following formula
$n=\dfrac{mass}{\text{molar mass}}$
\[n=\dfrac{W}{256}\]
Mass of benzene will be = $75 g$ \[=\dfrac{75}{1000}kg\] = $0.075 kg$
Now we can calculate molality of the solution, $m=\dfrac{moles}{\text{mass of solvent}}$
\[m=\dfrac{W}{256\times 0.075}\]= $19.2W molal$
The depression in the freezing point, \[\Delta {{T}_{f}}\] = $0.48K$
The depression in the freezing point constant = \[{{K}_{f}}\]= $5.12 Kg/mol$
\[\Delta {{T}_{f}}={{K}_{f}}m\]
Now, substituting these values in the given formula we get,
\[0.48=5.12\times \dfrac{W}{19.6}\]
\[W=\dfrac{0.48\times 19.2}{5.12}=1.8\]
Hence the correct answer is that $1.8 g$ of solute is to be dissolved.
Additional information:
Henry's law is defined as the gas laws formulated by William Henry in 1803 and it states that at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid. We can write it mathematically by stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid:
$C$ = $k[P_{gas}]$
Where, $C$ is the solubility of a gas at a fixed temperature in a particular solvent, $k$ is Henry's law constant and $[P_{gas}]$ is the partial pressure of the gas
Note: Freezing point can be defined as the temperature at which a liquid changes to a solid. The freezing point is lower than the melting point in the case of mixtures.
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