How mu Ch sucrose woulld you use to prepare 150 mL of a 15percentageew/v solutio

Question

How mu Ch sucrose woulld you use to prepare 150 mL of a 15percentageew/v solution? 2. How many moles of solute are there in 500 mL of a 2.4 M solution? 3. How would you make 250 mL of 2.0 M KC1? 4.You have a 2.5 M stock solution of NaCl and need to make 200 mL of a 0.5 M solution. How would you make it? (How mu Ch NaCi and how mu Ch water?) 5. Describe how to make 500 mL of a 4.0 M solution of KOH? 6. Ordinary household bleach in an aqueous solution of sodium hypochlorite, NaCIO. What is the molarity of bleach solution that contains 42.4 g of sodium hypochlorite, in a total volumee of 425 mL? b) express the concentration of the bleach solution in g/L c) express the concentration of the bleach solution in w/vpercentagee 7. Calculate each of the following quantities a) grams of solute in 255 mL of 0.045 M calcium acetate.

Explanation / Answer

1) we know that

%w/v = mass of sucrose x 100 / volume of solution

so

15 = mass of sucrose x 100 / 150

mass of sucrose = 22.5

so

22.5 grams of sucrose is needed

2) we know that

molarity = moles of solute x 1000 / volume (ml)

so

2.4 = moles of solute x 1000 / 500

moles of solute = 1.2

3)

moles of KCl = 2 x 250 / 1000

moles of KCl = 0.5

now

mass = moles x molar mass

so

mass of KCl = 0.5 x 74.55

mass of KCl = 37.275 g

take 37.275 g of KCl and add water upto a final volume of 250 ml

4)

we know that

for dilution

M1V1 = M2V2

so

2.5 x V1 = 200 x 0.5

V1 = 40

so

volume of stock solution = 40 ml

now

water added = 200 - 40 = 160 ml

so

160 ml of water should be added , no NaCl is required

5)

moles of KOH = 4 x 500 / 1000

moles of KCl = 2

now

mass = moles x molar mass

so

mass of KOH = 2 x 56

mass of KOH = 112 g

take 112 g of KOH and add water upto a final volume of 500 ml

6)

moles of NaCl0 = 42.4 / 74.44

moles of NaCl0 = 0.57

so

molarity = 0.57 x 1000 / 425

molarity = 1.34

b)

conc in g / L = 42.4 / 0.425

conc in g / L = 99.76

c) % w / v = 42.4 x 1000 / 425

% w / V = 9.976

7)

moles of calcium acetate = 0.045 x 255 / 1000

moles of calcium acetate = 0.011475

now

mass = moles x molar mass

so

mass of calcium acetate = 0.011475 x 158

so

mass of calcium acetate = 1.8 g

so

1.8 g of solute is present

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