Question

a. A soft drink contains 11.5% sucrose (C12H22O11) by mass. How much sucrose, in grams, is contained in 355 mL (12 oz) of the soft drink? (Assume a density of 1.04 g/mL.)

b.Calculate the molarity of a solution made by putting 55.8 g of NaNO3 into a beaker and diluting to 2.50 L.

c. Calculate the mass percent (m/m) of a solution prepared by dissolving 50.92 g of NaCl in 156.8 g of H 2 O .

Answer 1

mass % of glucose = 11.5 %
then mass of glucose = 11.5 g
mass of solution = 100 g
density of solution = 1.04 ml
volume of solution = mass/density
= 100g/1.04 g/ml
=96.1538 ml

96.1538 ml solution contain 11.5 g glucose
then 355 ml solution contain 11.5 g/96.1538 ml*355 ml
= 42.46 g
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mass of NaNO3 = 55.8 g
molar mass of NaNO3 = 84.99 g/mol
number of mols of NaNO3 =mass/molar mass
= 55.8 g/84.99 g/mol
volume of solution = 2.5 L
molarity = number of moles of solute/volume in Liters
= 55.8 g/84.99 g/mol*2.5 l
= 0.2626 mols/L
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mass of solute = 50.92 g
total mass of solution = 50.92 g+ 156.8 g
= 207.72 g

mass percent = mass of solute/mass of solution *100
=50.92 /207.72 *100
= 24.51 %
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all solved .Thank you