Chapter 14: Solutions 16 of 24 tem 16 Part B Constants 1 Periodic Table An aqueo

Question

Chapter 14: Solutions 16 of 24 tem 16 Part B Constants 1 Periodic Table An aqueous KNO, solution is made using 72.7g of KNOs diluted to a total solution volume of 2.03 Calculate the molality of the solution. (Assume a density of 1.06 g/ml for the solution.) Submit Previous Answers Request Answer incorrect; Try Again; 8 attempts remaining Not qute. Check through your calculations, you may have made a rounding error or used the wrong number of significant figures. Part C Calculate the mass percent of the solution. (Assume a density of 1.05 g/mL, for the solution.) mass percent . Submit Previous Answers Request Answer

Explanation / Answer

B)
volume of solution = 2.03 L = 2030 mL

mass of solution = density of solution * volume of solution
= 1.05 g/mL * 2030 mL
= 2131.5 g

mass of H2O = mass of solution - mass of KNO3
= 2131.5 g - 72.7 g
= 2058.5 g

Molar mass of KNO3,
MM = 1*MM(K) + 1*MM(N) + 3*MM(O)
= 1*39.1 + 1*14.01 + 3*16.0
= 101.11 g/mol


mass(KNO3)= 72.7 g

use:
number of mol of KNO3,
n = mass of KNO3/molar mass of KNO3
=(72.7 g)/(101.11 g/mol)
= 0.719 mol

m(solvent)= 2058.5 g
= 2.058 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.719 mol)/(2.0585 Kg)
= 0.3493 molal
Answer: 0.3493 molal

C)
mass percent = mass of KNO3 * 100 / mass of solution
= 72.7*100/2131.5
= 3.411 %
Answer: 3.411 %

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