The answer is enough solvent to make 1.00 L of solution. Since molarity is the number of moles of solute in one liter of solution, adding 0.500 mole solute to one liter solvent might not result to a solution with one liter total volume. Less than one liter solvent is first added to dissolve 0.500 mole solute and then the solution is carefully filled with more solvent until the solution reaches to one liter total volume. Hence, the resulting solution is a 0.500M concentration.
IF you are fat like your mom well i don't know
1 kg = 1000g
2.43 kg *1000g/1kg = 2430 g
H2SO4 + Na2CO3 → Na2SO4 + CO2 + H2O
The molarity of sulfuric acid if 1.78 L were used in the above reaction is
0.453 M (answer 2)
Calculation
find the moles of water produced = mass/molar mass
= 14.5 g /18 g/mol = 0.806 moles
by use of of mole ratio between H2So4 to H2O which is 1:1 the moles of H2SO4 is also = 0.806 moles
Molarity of H2SO4 is therefore = moles/volume in liters
= 0.806 mol/ 1.78 L = 0.453 M (answer 2)
For the answer to the question above, I can't help you directly because I don't have a calculator right now. But I'll show you how to solve this.
<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.
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<u>Mole solute</u> x mass of Water = Mol solute<u>
</u>kg Solvent
then
Mass of solute x <u> 1 </u> = molar mass
mole of solute
