how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C
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163 grams (3 sig. fig.).
<h3>Explanation</h3>- Formula of <em>carbon(IV) oxide</em> (a.k.a. carbon dioxide):
.
- Molar mass of
.
- Formula of ethyne: structural
or molecular
.
- Molar mass of
.
All carbon atoms in that 104 grams of ethyne will end up in . Number of moles of molecules in 104 grams of ethyne:
.
There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:
.
There are one carbon atom in each molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make
of
.
Mass of all those molecules:
. (3 sig. fig. as in the mass of ethyne.)
<u>Answer:</u> The freezing point of solution is -0.454°C
<u>Explanation:</u>
Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.
The equation used to calculate depression in freezing point follows:
To calculate the depression in freezing point, we use the equation:
Or,
where,
Freezing point of pure solution = 0°C
i = Vant hoff factor = 2
= molal freezing point elevation constant = 1.86°C/m
= Given mass of solute (KCl) = 5.0 g
= Molar mass of solute (KCl) = 74.55 g/mol
= Mass of solvent (water) = 550.0 g
Putting values in above equation, we get:
Hence, the freezing point of solution is -0.454°C