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.
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.
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.
Thank you sir
But i really don't understand your explanation in the first place
Yes it helped a lot
Thank you sir
To solve this problem, we need to use the concept of solubility and the given information about the saturation of the acid solution at different temperatures.
(a) Determining the mass of acid dissolved at 70°C:
Since we know that the residue of the acid after evaporation at 70°C had a mass of 5g, we can conclude that the initial mass of the acid in the solution was 25g (given). Therefore, the mass of acid dissolved at 70°C is 25g - 5g = 20g.
Determining the mass of acid dissolved at 40°C:
We need additional information to determine the mass of acid dissolved at 40°C. Let's calculate the number of moles of sodium hydroxide used in the neutralization reaction.
25g of the acid solution requires 50cm³ of 1M sodium hydroxide for neutralization. The concentration of a 1M solution is 1 mol/L, which means that 1L of the solution contains 1 mole of sodium hydroxide. Therefore, 1L of a 1M solution contains 1 mole or 1000mmol of sodium hydroxide.
The volume of 50cm³ is equal to 50/1000 = 0.05L. So, 0.05 L of the 1M sodium hydroxide solution contains 0.05 moles or 50 mmol of sodium hydroxide.
According to the neutralization reaction (H2X + 2NaOH = Na2X + 2H2O), 2 moles of sodium hydroxide react with 1 mole of acid (H2X). Therefore, the number of moles of acid in the 25g solution at 40°C is 50 mmol / 2 = 25 mmol.
Now, we need to find the mass of 25 mmol of acid. The molar mass of the acid is required for this calculation.
(b) Plotting the graph:
To plot a graph of the number of grams of acid dissolving in 100g of water against the temperature, you need to have data points for various temperatures. Unfortunately, the problem only provides information for 70°C and 40°C. Therefore, we can only plot two data points on the graph.
The x-axis will represent the temperature, with a scale of 2cm for 10°C. So, the two data points will be at 70°C and 40°C.
The y-axis will represent the mass of acid dissolving in 100g of water, with a scale of 2cm for 5g. The two data points are 20g at 70°C and the mass we still need to calculate for 40°C.
(c) Predicting the maximum mass of acid at 0°C:
Since we only have two data points at 70°C and 40°C, we can draw a straight line connecting these two points. Then, we can extend the line to intersect with the x-axis (temperature axis) at 0°C. The corresponding y-axis (mass of acid) value on the x-axis will give us the predicted maximum mass of acid that can dissolve in 100g of water at 0°C.