Correct option is : 309 g
Explanation
Given : mass of salt = 40 g
mass of solution = 269 g
mass of solution = (mass of salt) + (mass of solution)
mass of solution = (40 g) + (269 g)
mass of solution = 309 g
A generic salt, AB3, has a molar mass of 269 g/mol and a solubility of 2.10 g/L at 25 °C. What is the Ksp of this salt at 25 °C? AB3(s)−⇀↽−A3+(aq)+3B−(aq)
A 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. What is the solubility (in g salt/100 g of water) of the salt? (2 points) How much water would it take to dissolve 25 g of this salt? (2 points) If 10.0 g of this salt is mixed with 15.0 g of water, what percentage of the salt dissolves? (2 points)
We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 41.1 mL of the NaOH solution to reach the equivalence point.What is the molar mass of HA and What is the pKa value of HA(aq)?
1) What mass of water is needed to dissolve 43.5 g of copper(II) chloride to prepare a 0.521 molality solution? 2) Calculate the boiling point of a solution made from 224 g of magnesium chloride dissolved in 625 g of water. Kb=0.512 °C/m.
We dissolve 2.77 g of an unknown acid, HA, in enough water to
produce 25.0 mL of solution. The pH of this solution of HA(aq) is
1.33. We titrate this solution with a 0.250 M solution of NaOH. It
takes 41.9 mL of the NaOH solution to reach the equivalence
point.
(a) (2 points) What is the molar mass of HA?
(b) (2 points) What is the pKa value of HA(aq)?
(c) (2 points) What is the pH at the...
We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 44.5 mL of the NaOH solution to reach the equivalence point. (a) What is the molar mass of HA? (b) What is the pKa value of HA(aq)? (c) What is the pH at the equivalence point? (d) What is the...
We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 41.8 mL of the NaOH solution to reach the equivalence point. (a) What is the molar mass of HA? (b) What is the pKa value of HA(aq)? (c) What is the pH at the equivalence point? (d) What is the...
We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 41.6 mL of the NaOH solution to reach the equivalence point. (a) What is the molar mass of HA? (b) What is the pKa value of HA(aq)? (c) What is the pH at the equivalence point? (d) What is the...
We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 41.4 mL of the NaOH solution to reach the equivalence point. (a) What is the molar mass of HA? (b) What is the pKa value of HA(aq)? (c) What is the pH at the equivalence point? (d) What is the...
We dissolve 2.77 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. The pH of this solution of HA(aq) is 1.33. We titrate this solution with a 0.250 M solution of NaOH. It takes 43.3 mL of the NaOH solution to reach the equivalence point. (a)What is the molar mass of HA? (b)What is the pKa value of HA(aq)? (c)What is the pH at the equivalence point? (d) What is the pH one third...