
How much NaCl should be added to 600 g of water (\[\rho \]= 1.00 g/mL) to decrease the freezing point of water to \[ - 0.2^\circ C\] ? (The freezing point depression constant for water = 2 K Kg \[mo{l^{ - 1}}\])
Answer
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Hint: The normal freezing point of water is \[0^\circ C\] i.e. at this temperature, water transforms into its solid state of ice. When common table salt is added to water, salt molecules block water molecules from packing together when the temperature is lowered to \[0^\circ C\]. It then prevents the molecules of water from becoming ice. More water molecules leave the solid (ice) phase than the ones entering the solid phase. Freezing point depression (lowering of the freezing point) occurs when the freezing point of the liquid is lowered by the addition of solute.
Complete Step by Step Solution:
NaCl is strong electrolyte and gives 2 ions in the solution i.e. \[N{a^ + }\] and \[C{l^ - }\] . Hence, i = 2.
Molarity = \[\dfrac{{w \times 1000}}{{58.5 \times 600}}\]
\[\Delta {T_f}\] = \[0.2^\circ C\] (Change in freezing point after the addition of NaCl)
\[\Delta {T_f}\] = \[i \times {k_f} \times m\] (\[{k_f}\] = Freezing point depression constant. )
On solving,
\[\Delta {T_f}\]= 0.2 K ; m = \[\dfrac{{\Delta {T_f}}}{{i \times \Delta {K_f}}} = \dfrac{{0.2}}{{2 \times 2}}\]= 0.05 m
For Nacl, i=2
0.05 moles of NaCl in 1000g of \[{H_2}O\]
So, for 600g of \[{H_2}O\] = \[(\dfrac{{0.05}}{{1000}}) \times 600\] = 0.03 moles of NaCl = \[0.03 \times 58.5g\] of NaCl = 1.76g of NaCl
Hence, w = 1.76 grams
Additional Information:Water is an unusual liquid. Like every other liquid, water tends to decrease in volume when cooled but up to a certain temperature i.e. \[4^\circ C\]. After that, any temperature lower would lead to an expansion of its volume.
Note: Salt makes ice colder as it prevents melted water from freezing all over again. Melting is an endothermic process, so it lowers the temperature. On sidewalks and roadways, salt aids in the melting of ice and the prevention of it refreezing. On the other hand, adding salt to ice makes it colder so it helps in freezing ice cream.
Complete Step by Step Solution:
NaCl is strong electrolyte and gives 2 ions in the solution i.e. \[N{a^ + }\] and \[C{l^ - }\] . Hence, i = 2.
Molarity = \[\dfrac{{w \times 1000}}{{58.5 \times 600}}\]
\[\Delta {T_f}\] = \[0.2^\circ C\] (Change in freezing point after the addition of NaCl)
\[\Delta {T_f}\] = \[i \times {k_f} \times m\] (\[{k_f}\] = Freezing point depression constant. )
On solving,
\[\Delta {T_f}\]= 0.2 K ; m = \[\dfrac{{\Delta {T_f}}}{{i \times \Delta {K_f}}} = \dfrac{{0.2}}{{2 \times 2}}\]= 0.05 m
For Nacl, i=2
0.05 moles of NaCl in 1000g of \[{H_2}O\]
So, for 600g of \[{H_2}O\] = \[(\dfrac{{0.05}}{{1000}}) \times 600\] = 0.03 moles of NaCl = \[0.03 \times 58.5g\] of NaCl = 1.76g of NaCl
Hence, w = 1.76 grams
Additional Information:Water is an unusual liquid. Like every other liquid, water tends to decrease in volume when cooled but up to a certain temperature i.e. \[4^\circ C\]. After that, any temperature lower would lead to an expansion of its volume.
Note: Salt makes ice colder as it prevents melted water from freezing all over again. Melting is an endothermic process, so it lowers the temperature. On sidewalks and roadways, salt aids in the melting of ice and the prevention of it refreezing. On the other hand, adding salt to ice makes it colder so it helps in freezing ice cream.
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