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lorasvet [3.4K]
3 years ago
11

approximately how many grams of sodium chloride (NaCl) are required to prepare 500. mL of 3.00 M solution

Chemistry
1 answer:
WITCHER [35]3 years ago
4 0
3.00 M  means 3.00 mol/L. So if you want 500 mL=0.5 L, it needs 3.00 mol/L * 0.5 L = 1.50 mol. For NaCl, the molar mass is 23+35.5=58.5 g/mol. So the amount of NaCl needed is 1.50*58.5=87.75 g. So the answer is 87.75 grams.
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<span>use the freezing point depression formula for this one: delta T = i * m * K where K is a constant, m is the molality (mol solute/kg solvent), and i is the van'hoff factor the van hoff factor is the number of ions that your salt dissociates into. Since it's an ALKALI flouride salt, how many ions? k is just a constant, you get it from a table in your textbook somewhere So you have everything to solve for the molality of the solution, once you did that, multiplying it by the mass of water to find the mols of the salt. Take the mass of the salt and divide by this mols to figure out the molar mass, and then compare it with the periodic table to identify the salt.
</span>
<u>Mole solute</u>   x  mass of Water  = Mol solute<u>
</u>kg Solvent

then 

Mass of solute  x  <u>            1             </u>   =  molar mass
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