Problem 9.48 - Enhanced- with Feedback Part A 2.80 L of a 1.60 M NaOH solution E

Question

Problem 9.48 - Enhanced- with Feedback Part A 2.80 L of a 1.60 M NaOH solution Express your answer with the appropriate units. Constants i Perk deTabl Calculate the grams following. of solute needed to prepare each of the You may want to reference (Pages 298-305) Section 9.4 while completing this problem. Value Units Submit PartB 4.00 L of a 0.130 M CaCl2 solution Express your answer with the appropriate units. Value Units Submit PartC 225 mL of a 2.45 M NaNOs solution Express your answer with the appropriate units. Value Units

Explanation / Answer

A)
volume , V = 2.80 L


use:
number of mol,
n = Molarity * Volume
= 1.6*2.8
= 4.48 mol

Molar mass of NaOH,
MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol

use:
mass of NaOH,
m = number of mol * molar mass
= 4.48 mol * 39.998 g/mol
= 179 g
Answer: 179 g

B)
volume , V = 4.00 L


use:
number of mol,
n = Molarity * Volume
= 0.13*4
= 0.52 mol

Molar mass of CaCl2,
MM = 1*MM(Ca) + 2*MM(Cl)
= 1*40.08 + 2*35.45
= 110.98 g/mol

use:
mass of CaCl2,
m = number of mol * molar mass
= 0.52 mol * 110.98 g/mol
= 57.7 g
Answer: 57.7 g

C)
volume , V = 225 mL
= 0.225 L


use:
number of mol,
n = Molarity * Volume
= 2.45*0.225
= 0.5513 mol

Molar mass of NaNO3,
MM = 1*MM(Na) + 1*MM(N) + 3*MM(O)
= 1*22.99 + 1*14.01 + 3*16.0
= 85 g/mol

use:
mass of NaNO3,
m = number of mol * molar mass
= 0.5513 mol * 85 g/mol
= 46.9 g
Answer: 46.9 g

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