Question

Explain how you would prepare the following solutions using pure solute and water. Assume water has a density of 1.00 g/mL.

a) 200mL of 0.150M NA2SO4 solution

b) 250 mL of 0.250 M Zn (NO3)2 solution

c) 150 g of 2.25% (w/w) NaCl solution

d) 125 mL of 0.75% (w/v) KCl solution

Please show step by step and in a simple way so I understand. Thank you so much.

Answer 1

a)we have a molarity formula as follows

Molarity = (W /MW) (1000 / V in mL)

Molarity = 0.150 M

W = ?

MW = 142.04 g/mol

V = 200 mL

substitute these values in above formula

0.150 = (W / 142.04) (1000 / 200)

0.03 = W / 142.04

W = 4.26 g

4.26 g Na2SO4 dissolve in 200 mL solution

b) same as part 1 use molarity formula

Molarity = (W /MW) (1000 / V in mL)

Molarity = 0.250 M

MW = 189.36 g/mol

W = ?

V = 250 mL

substitute these values in above formula

0.250 = (W / 189.36) (1000 / 250)

0.0625 = W / 189.36

W = 11.84 g

11.84 g Zn(NO3)2 dissolve in 250 mL solution

c)

we have a formula for % mass as follows

% w/w = (weight of solute / weight of solution) x 100

% w/w = 2.25

weight of solution = 150 g

weight of solute = ?

substitute these values in above formula

2.25 = (weight of solut / 150) x 100

0.0225 = weight of solute / 150

weight of solute = 3.375 g

3.375 g NaCl dissolve in (150 - 3.37) 146.625 g H2O

d)

we have a formula for % m/v

% w/v = (weight of solute / volume of solution) x 100

% m/v = 0.75

volume of solution = 125 mL

0.75 = (weight of solute / 125) x 100

weight of solute / 125 = 0.0075

weight of solute = 0.94 g

0.94 g KCl dissolve in 125 mL solution