1)What mass of an aqueous 21.4% potassium chloride solution contains 86.4 g of w

Question

1)What mass of an aqueous 21.4% potassium chloride solution contains 86.4 g of water?

2)A concentrated sulfuric acid solution is 96.0% H2SO4 by mass and has a density of 1.84 g/mL at 25°C. What is the molarity of H2SO4?

3)How many moles of urea (60. g/mol) must be dissolved in 63.3 g of water to give a 3.2 m solution?

4)When a 43.9-g sample of an unknown compound is dissolved in 500. g of benzene, the freezing point of the resulting solution is 3.77°C. The freezing point of pure benzene is 5.48°C, and Kf for benzene is 5.12°C/m. Calculate the molar mass of the unknown compound.

5)The following reaction is investigated (assume an ideal gas mixture):

Initially there are 0.100 mol of N2O and 0.25 mol of N2H4, in a 10.0-L container. If there are 0.061 mol of N2O at equilibrium, how many moles of N2 are present at equilibrium?

6)

In the reaction

CaO(s) + CO2(g) CaCO3(s),

A.CaO is the Lewis acid and CaCO3 is its conjugate base.

B.O2– acts as a Lewis base and CO2 acts as a Lewis acid.

C.Ca2+ acts as a Lewis acid and CO32– acts as a Lewis base.

D.O2– acts as a Lewis base and Ca2+ acts as a Lewis acid.

E.CO2 is the Lewis acid and CaCO3 is its conjugate base.

7)Rank the following in order of decreasing acid strength in aqueous solution: HCl, HOCl, HOBr, HOI.

8)

At 25°C a solution has a hydroxide-ion concentration of 4.91 × 10-5M. What is its hydronium-ion concentration?

9)

A 0.20 M solution of a weak monoprotic acid is 0.22 % ionized. What is the acid-ionization constant, Ka, for this acid?

10)A 0.10 M aqueous solution of a weak acid HA has a pH of 6.00. What is the value of Ka for HA?

11)

What is the pH of a solution that is formed at 25°C by combining 600 mL of 0.020 M NaOH with 300 mL of 0.040 M HCl?

12)What is the hydronium-ion concentration of a solution formed by combining 700. mL of 0.20 M HCl with 300. mL of 0.53 M NaOH at 25°C?

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Explanation / Answer

1.

21.4% KCl means 21.4g of KCl is present in 100ml of water.

In case of water 1g = 1ml so 86.4g = 86.4ml

So amount of KCl in 80ml of water = 86.4*221.4/100
= 18.48g.

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