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icang [17]
3 years ago
14

A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti

on
Chemistry
2 answers:
IceJOKER [234]3 years ago
6 0

Answer : The molarity and molality of the solution is, 1.00 mole/L and 0.904 mole/Kg respectively.

Solution : Given,

Density of solution = 1.19 g/ml

Molar mass of sodium carbonate (solute) = 84 g/mole

7.06% aqueous solution of sodium bicarbonate means that 7.06 gram of sodium bicarbonate is present in 100 g of solution.

Mass of sodium bicarbonate (solute) = 7.06 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 7.06 = 92.94 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.19g/ml}=84ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Moles of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{7.06g\times 1000}{84g/mole\times 84ml}=1.00mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Moles of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{7.06g\times 1000}{84g/mole\times 92.94g}=0.904mole/Kg

Therefore, the molarity and molality of the solution is, 1.00 mole/L and 0.904 mole/Kg respectively.

lilavasa [31]3 years ago
5 0

<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg

<em>Molar concentration </em>

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

<em>c</em> = 1.14 mol/1 L = 1.14 mol/L

<em>Molal concentration</em>

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg

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