Answer:
Solution of isopropanol is 10.25 molal
Explanation:
615 g of isopropanol (C3H7OH) per liter
We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.
Density of water 1g/mL → Density = Mass of water / 1000 mL of water
Notice we converted the L to mL
Mass of water = 1000 g (which is the same to say 1kg)
Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles
Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m
The answer & explanation for this question is given in the attachment below.
Remember: heat lost = heat gained
When calculating heat loss or gain, remember
mass*(spec heat cap)*(change in T)
The unknown loses heat- we don't know the spec heat cap, so we'll call it x.
The water gains. I've omitted the units, but always use when solving problems on your own.
75*x*(96.5-37.1) = 1150*4.184*(37.1-25)
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Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC
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This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:
H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O
So, we do the stoichiometric calculations. The important data we should know is the molar mass of oxalic acid which is equal to 90 g/mol.
(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%
This is impossible. The purity can't be more than 100%. Looking at our calculations and the balance reaction, all steps were done correctly. So, I think there is some typographical error in the given. The mass of the sample should be 7.984 g. Then, the answer would be 48.87% purity.