Concentration of Solutions Describe how you would prepare 250 mL of a 0.707 M Na
ID: 949663 • Letter: C
Question
Concentration of Solutions Describe how you would prepare 250 mL of a 0.707 M NaNO_3 solution. How many grams of KOH are present in 35.0 mL of a 5.50 M solution? Calculate the molarity of each of the following solution (a) 29.0g of ethanol (C_2H_5OH) in 545 mL of solution, (b)15.4 of sucrose (C_12H_22O_11) in 74.0 mL of solution, (c) 9.00g of sodium chloride (NaCI) in 86.4mL of solution. Calculate the volume in mL of a solution required to provide the following: (a) 2.14g of sodium chloride from a 0.270 M solution (b) 4.30 g of ethanol from a 1.50 M solution, (c) 0.85 g of acetic acid (CH_3COOH) from a 0.30 M solution. What volume of of 0.416 M MG (NO_3)_2 should be added to 255 mL of 0.102 M KNO_3 to produce a solution with a concentration of 0.278 M^NO_3 ions? Assume volumes are additive.Explanation / Answer
4.62)
we know that
moles = molarity x volume / 1000
so
moles of NaN03 = 0.707 x 250 / 1000
moles of NaN03 = 176.75 x 10-3
now
we know that
mass = moles x molar mass
so
mass of NaN03 = 176.75 x 10-3 x 85
mass of NaN03 = 15.02375
now
so the steps are
a) take 15.02375 g of NaN03
b) add water to make the final volume to 250 ml
4.69)
let V ml of Mg(N03)2 be added
now
we know that
moles = molarity x volume (ml) / 1000
so
moles of Mg(N03)2 = 0.416 x V / 1000 = 0.416V x 10-3
so
moles of N03- = 2 x 0.416 x V x 10-3 = 0.832V x 10-3
now
moles of KN03 = 0.102 x 255 x 10-3 = 26.01 x 10-3
so
moles of N03- = 26.01 x 10-3
now
total moles of N03- = 0.832V x 10-3 + 26.01 x 10-3
total moles of N03- = ( 0.832V + 26.01) x 10-3
now
final volume = ( 255 + V) ml
now
molarity = moles x 1000 / volume (ml)
so
0.278 = (0.832V + 26.01) x 10-3 x 1000 / (255 + V)
70.89 + 0.278V = 0.832V + 26.01
V = 81
so
81 ml of Mg(N03)2 should be added
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