A solution of sodium bicarbonate is prepared by adding 45.00g of sodium bicarbonate to a 1.00-L volumetric flask and adding distilled water until it reaches the 1.00-L mark. Calculate the concentration of sodium in units of:

a) milligrams per liter
b) molarity
c) normality
d) milligrams per liter as CaCO3

sodium bicarbonate is NaHCO3.

45.00 g/L. Convert to mg/mL. To convert g to mg, multiply by 1000. To convert from L t mL, multiply by 1000. So 45.00 g/L is
45.00 g/L x (1000 mg/g)(1 L/1000 mL) = 45.00 mg/mL for NaHCO3. Since there is 1 atom Na per molecule of NaHCO3, that is the concn of Na^+ in mg/mL.

b. Convert g NaHCO3 to moles. That will be moles Na^+ also. Then moles/L = M.

c. There is 1 H in NaHCO3, therefore, N = M.

d. Convert 45.00 g NaHCO3/L to g CaCO3/L, then change to mg/L.
Post your work if you get stuck.

a) 45.00g/L(1000mg/g)(1L/1000mL)=45mg/mL

b)NaHCO3: 22.99+1+12+3(16)= 83.99g/mol
prop. of Na (22.99/83.99)=0.54
c)?
d)?

a is ok.

b you have close to the correct answer but not the correct procedure.
g NaHCO3 = 45.00 molar mass NaHCO3 = 84.007 according to my net source.
moles NaHCO3 = (45.00/84.007) = 0.535669 and since you have 4 significant figures in the 45.00 g NaHCO3, this should be rounded to 0.5357 moles NaHCO3. Since there is 1 mole Na^+ in a mole of NaHCO3, the moles Na^+ = 0.5357 and that is the molarity. (22.99/83.99 is NOT 0.54--nor close to that).

c. N = M x #H atoms. N = 0.5357 x 1 = ??
d. 45.00 g NaHCO3 x (molar mass CaCO3/molar mass NaHCO3) = 45.00 g NaHCO3 x(100.087/84.007) = ??g CaCO3/L
Change that to mg CaCO3/L.

In my book the anser for D is 2.68 x 10^4 mg/L as CaCO3, my first thought was to do what you did but i cant get to the answer in the book, am I missing something very obvious here? Im getting like 5.4 x 10^4

why is the molar mass 50 not 100 for the caco3?

a) Sure, I can calculate that for you. Sodium bicarbonate has a molar mass of approximately 84 g/mol. Since you added 45.00g to a total volume of 1.00 L, the concentration of sodium bicarbonate in milligrams per liter (mg/L) would be 45.00 g/L or 45,000 mg/L.

b) To calculate the molarity, we need to convert the grams of sodium bicarbonate to moles and then divide by the volume in liters. So, the number of moles of sodium bicarbonate is 45.00 g / 84.00 g/mol = 0.54 mol. Hence, the molarity is 0.54 mol/1 L or 0.54 M.

c) Normality is a unit that takes into account the number of equivalents of the solute. In the case of sodium bicarbonate, it has one sodium ion and one bicarbonate ion, so its normality would be the same as its molarity, which is 0.54 N.

d) To calculate the concentration in milligrams per liter as CaCO3, we need to consider the molecular weight of sodium bicarbonate and its ratio to CaCO3. The molecular weight of sodium bicarbonate is 84 g/mol, while the molecular weight of CaCO3 is 100 g/mol. Therefore, the concentration in mg/L as CaCO3 would be 45.00 g/L × (100 g/mol / 84 g/mol) = 53.57 mg/L as CaCO3.

I hope these answers help, but if they don't make you smile, I can always try again!

To calculate the concentration of sodium in various units, we need to determine the number of moles of sodium bicarbonate in the solution.

a) Milligrams per liter (mg/L):
To calculate the concentration in milligrams per liter, we first need to find the number of moles of sodium bicarbonate using its molar mass.

The molar mass of sodium bicarbonate (NaHCO3) =
(1 × atomic mass of Na) + (1 × atomic mass of H) + (3 × atomic mass of C) + (3 × atomic mass of O)

= (1 × 22.99 g/mol) + (1 × 1.01 g/mol) + (3 × 12.01 g/mol) + (3 × 16.00 g/mol)
= 84.01 g/mol

Now, we can calculate the number of moles using the given mass:
Number of moles of NaHCO3 = mass (g) / molar mass (g/mol)
= 45.00 g / 84.01 g/mol

Once we have the number of moles of NaHCO3, we can convert it to milligrams:
Concentration (mg/L) = (moles / volume in liters) × (1000 mg/g)

b) Molarity (M):
To calculate the molarity, we divide the number of moles of sodium bicarbonate by the volume in liters.

Since the solution is prepared in a 1.00-L volumetric flask, the volume is equal to 1.00 L.

Concentration (M) = moles / volume in liters

c) Normality (N):
Normality is a measure of the number of equivalents of a solute in a liter of solution. In this case, we can consider sodium bicarbonate (NaHCO3) to be a monoprotic acid.

Normality (N) = moles of solute / volume in liters

d) Milligrams per liter as CaCO3:
To calculate the concentration in milligrams per liter as CaCO3, we need to consider the molar mass of calcium carbonate (CaCO3), which is 100.09 g/mol.

Number of moles of CaCO3 = Number of moles of NaHCO3
Concentration (mg/L as CaCO3) = (moles of CaCO3 / volume in liters) × (molar mass of CaCO3 in mg/mol)

Note: For parts a), c), and d), we use the number of moles determined in the first step.

Once you have the number of moles of sodium bicarbonate in the solution, you can use these equations to calculate the concentration in different units.