I'm mostly unable to help because there just isn't enough information given.
a. mass acid dissolved in 25 g of the saturated solutioin @ 70o C is 5 grams in the problem. You don't know and can't calculate how much dissolves @40o C If it take 50 cc of 1 molar NaOH that is 50 millimolar NaOH or 1/2 that or 25 mmolar H2X BUT without knowing the molar mass of the H2X the solubility in grams is unknown.
b. You don't have enough data (can't determine mass acid) AND we can't draw diagrams/graphs on this site.
c. Can't do graphs.
25g of saturated solution of an acid in water at 70°c were evaporated to dryness.The residue of the acid had a mass of 5g
25g of saturated solution of the acid in water at 40°c needed 50cm³ of 1M solution of sodium hydroxide for complete neutralisation reaction according to
H2X +2NaOH =Na2X +2H2O
(a) Determine the mass of acid which would dissolve in water at 70°c and 40°c
(b) plot a graph of the number of grammes of acid ,dissolving in 100g of water against the temperature.Use a scale of 2cm for 10°C along x axis and 2cm for 5g along y axis.
(c) use your graph to predict the maximum mass of acid which will dissolve in 100g of water at 0°C.
4 answers
Thank you sir
But i really don't understand your explanation in the first place
But i really don't understand your explanation in the first place
OK. Here we go a second try.
Parts 2 and 3 we can forget about since this site does not allow us to draw diagrams or graphs or pictures or anything of that order.
For #1, the problem states that the residue from the 25 g sample of the saturated solution had a mass of 5 g and that was for the 70 C sample as stated. For the 40 C sample, the problem gives you a method to determine the mass. Unfortunately they didn't give enough information. They told you that a sample of the acid (25 g of the saturated solution) was neutralized by 50 cc of 1 M NaOH and gave you the equation of
H2X + 2NaOH ==> Na2X + 2H2O.
So how many moles of NaOH were used? mols = M x L = 1 M x 0.050 = 0.050 mols of NaOH. From the equation you can see that 1 mol H2X reacts exactly with 2 mols NaOH; therefore mols H2X titrated in that 25 g sample @ 40 C was 1/2 * 0.050 = 0.025 mols (or 25 millimoles that I used earlier).
Now IF we knew the molar mass we could use grams = molar mass x moles BUT we don't have that so we CAN calculate the moles but we don't have the molar mass. So the only part of the question that can be answered is that the solubility @ 70 C is 5 g because that's what the problem said. I hope this helps.
Parts 2 and 3 we can forget about since this site does not allow us to draw diagrams or graphs or pictures or anything of that order.
For #1, the problem states that the residue from the 25 g sample of the saturated solution had a mass of 5 g and that was for the 70 C sample as stated. For the 40 C sample, the problem gives you a method to determine the mass. Unfortunately they didn't give enough information. They told you that a sample of the acid (25 g of the saturated solution) was neutralized by 50 cc of 1 M NaOH and gave you the equation of
H2X + 2NaOH ==> Na2X + 2H2O.
So how many moles of NaOH were used? mols = M x L = 1 M x 0.050 = 0.050 mols of NaOH. From the equation you can see that 1 mol H2X reacts exactly with 2 mols NaOH; therefore mols H2X titrated in that 25 g sample @ 40 C was 1/2 * 0.050 = 0.025 mols (or 25 millimoles that I used earlier).
Now IF we knew the molar mass we could use grams = molar mass x moles BUT we don't have that so we CAN calculate the moles but we don't have the molar mass. So the only part of the question that can be answered is that the solubility @ 70 C is 5 g because that's what the problem said. I hope this helps.
Yes it helped a lot
Thank you sir
Thank you sir
1 answer