What mass of sodium sulfite would be required to form 500.0 mL of a solution wit

Question

What mass of sodium sulfite would be required to form 500.0 mL of a solution with [Na+] = 0.223 M?

2. Assuming the final density of the solution in Question

1 is 1.032 g/mL, calculate the following sodium sulfite

in the solution

mass percent

molality.

3. What would be the ppm Na+ if 2.00 L of water

were added to the solution in Question 1?

What mass of sodium sulfite would be required to form 500.0 mL of a solution with [Na+] = 0.223 M?

2. Assuming the final density of the solution in Question

1 is 1.032 g/mL, calculate the following sodium sulfite

in the solution

mass percent

molality.

3. What would be the ppm Na+ if 2.00 L of water

were added to the solution in Question 1?

Whaitl mass of sodium sulfite would be roquirod to fom $00 0mL of a solution with [Na1-0.223 M? 2 Assuming the final density of the solution in Qt1 is 1.032 gml calculate the llowing for sodium sulfite in the solution molality What would be the prom Nat i2.00 L of water ww added to the solution in Question 17 4 Estimate the freezing point of the following agaeous sollutions: Note: K, (H,0)-186 Cm 0.10m COH1206a) b) a saturated solution of Li2C03, with solubility 158g1000 g water at OeC When you cook spaghetti, you sometimes add a pich of sallh to the boiling water Estimate the boiling point of 2.00 quarts of water to which 2.0oaNaCT) ve been aded. Kb 01,0) 051 Cm

Explanation / Answer

Solution:- (1) Sodium sulfite is Na2SO3. The equation for it's dissociation to give the ions is written as...

Na2SO3(aq) ------> 2Na+(aq) + SO32-(aq)

From this equation, 1 mol of the salt gives two moles of Na+.

500.0 mL x (1L/1000mL) x (0.223 mol Na+/1L) x (1mol Na2SO3/2 mol Na+) x (126 g Na2SO3/1mol Na2SO3)

= 7.02 g Na2SO3

(2) Volume of solution = 500.0 mL

density of solution = 1.032 g/mL

mass = volume x density

500.0 mL x (1.032 g/mL) = 516.0 g

So, mass of solution is 516.0 g and the mass of solute that is sodium sulfite is 7.02 g.

Mass percent of solute = (mass of solute/mass of solution)*100

mass percent of sodium sulfite = (7.02/516.0)*100 = 1.36%

molality is the moles of solute per kg of solvent.

mass of solvent = 516.0 g - 7.02 g = 508.98 g = 0.509 kg

Let's convert grams of solute to moles.

7.02 g x (1mol Na2SO3/.126g) = 0.0557 mol

So, molality = 0.0557 mol/0.509 kg = 0.109m

(3) Total volume of Solution = 2.5 L

let's calculate the grams of Na+ present in the solution.

500.0 mL x (1L/1000mL) x (0.223 mol/L) x (22.99g/1mol) = 2.56 g Na+

ppm is mg of solute per liter of solution, Here we have 2.56 g that is 2560 mg of Na+ present in 2.5 L solution,

so, ppm concentration of Na+ = 2560 mg/2.5L = 1024ppm

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