5. A solution contains 2.0mg of NaCl, 6.00g of glucose (C&H12O6), and 112.0g of H,O. a. What is the molality (m) of the glucose? mak saute kg solvant 2.09 = 0.002, Nael 0.033 mals CaHiabe %30.29M CaHILOL 0.112 kg H20 Imal 80.14g 2 0.033 mols Cat20 12.6g H20 b. What is the concentration of the Na* in parts per million (ppm)? 35.45 gul 0.G07 0.002 g Naci. I mat 0.0012143Nat 35.458 0.6606342 nd
5. A solution contains 2.0mg of NaCl, 6.00g of glucose (C&H12O6), and 112.0g of H,O. a. What is the molality (m) of the glucose? mak saute kg solvant 2.09 = 0.002, Nael 0.033 mals CaHiabe %30.29M CaHILOL 0.112 kg H20 Imal 80.14g 2 0.033 mols Cat20 12.6g H20 b. What is the concentration of the Na* in parts per million (ppm)? 35.45 gul 0.G07 0.002 g Naci. I mat 0.0012143Nat 35.458 0.6606342 nd
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Not sure if (a.) is correct. But I need help understanding how to do (b.).

Transcribed Image Text:d. Mole Fraction (X) of the HCI
5. A solution contains 2.0mg of NaCl, 6.00g of glucose (C,H12O6), and 112.0g
of H2O.
Na CL
2 35:45
22.99
a. What is the molality (m) of the glucose?
2.09
= 0.002, Nael
mak sdute
%3D
0.033 mals CaHiale
=0.29m CabHizOy
kg solvant
0.112 kg H20
Imala 0.033 mols Catr2.Che
780 6g
L =0.112 kg Ha0
12.61(G).
10009
b. What is the concentration of the Na" in parts per million (ppm)?
186.16 glmed
d.002 g Naci
X a.407
35.45
H2O
4 0.G07
I mat
6,001214& Nat
=0.6606342 ned
18.02 smol
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