1. Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56g
ID: 815604 • Letter: 1
Question
1. Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56g of NaCl in 172.3g of H2O.
Express your answer to four significant figures.
2. Vinegar is a solution of acetic acid in water. If a 185mL bottle of distilled vinegar contains 26.4mL of acetic acid, what is the volume percent (v/v) of the solution?
Express your answer to three significant figures.
3. Calculate the mass/volume percent (m/v) of 12.5g NaCl in 74.0mL of solution.
Express your answer to three significant figures.
4. Calculate the molarity (M) of 152.1g of H2SO4 in 1.285L of solution.
Express your answer to four significant figures.
6. How many mL of a 0.45{ m %} (m/v) KCl solution can be prepared from 500mL of a 8.5{ m %} (m/v) stock solution?
Express your answer using two significant figures.
0.075 %v/v 7.5 %v/v 7.0 %v/v 0.070 %v/v 1. Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56g of NaCl in 172.3g of H2O. Express your answer to four significant figures. 2. Vinegar is a solution of acetic acid in water. If a 185mL bottle of distilled vinegar contains 26.4mL of acetic acid, what is the volume percent (v/v) of the solution? Express your answer to three significant figures. 3. Calculate the mass/volume percent (m/v) of 12.5g NaCl in 74.0mL of solution. Express your answer to three significant figures. 4. Calculate the molarity (M) of 152.1g of H2SO4 in 1.285L of solution. Express your answer to four significant figures. 5. What is the %v/v concentration of a sample of wine that contains 15 mL of ethyl alcohol in 200 mL of wine? 0.075 %v/v 7.5 %v/v 7.0 %v/v 0.070 %v/v 6. How many mL of a 0.45{rm %} (m/v) KCl solution can be prepared from 500mL of a 8.5{rm %} (m/v) stock solution? Express your answer using two significant figures.Explanation / Answer
A) The formula we are gonna use is % = mass of salt / total mass
the total mass is the mass of salt + mass of water
total mass = 51.56 g of NaCl + 172.3g of Water = 223.86 g
% = 51.56/223.86 = 0.2303 * 100% = 23.03%
% = 23.03%
b) Total volume = 185 mL, of that 26.4 mL is acetica acid
The volume percent is calculated as % = volume of substance / total volume * 100%
26.4 mL acetic acid /185 mL vinegar * 100 % = 14.3%
c) m/v percent is interesting one
% = mass of solute (which is NaCl) / volume of solvent (which is water or the solutoin) * 100%
% = 12.5 g / 74 ml *100% = 16.9%
d) The difinition of Molarity is = moles per unit volume (L)
M = moles / Volume (in L)
MW = H2SO4 = 98 g/gmol
moles of H2SO4 in 152.1 g of H2SO4 = mass/MW
moles = 152.1g / 98 g/gmol = 1.55 moles of H2SO4
M = 1.55 mol / 1.285 L = 1.206 M
e) The v/v concentraion of wine with 15 mL of ethanol in 200 ml of wine
% v/v = volume of substance / total volume of solution *100% = 15 ml / 200 ml * 100%
% v/v = 7.5 %
f) 0.45% m/v of KCl solution with 500 ml of 8.5% m/v soltion
M1V1 = C1
M2V2 = C2
500 ml *8.5% = 0.45% * V
V = 500 * (8.5/.45)
V = 9444.44 ml
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