1. A solution contains 35 parts per thousand (g kg -1 ) NaCl. Calculate the conc

Question

1. A solution contains 35 parts per thousand (g kg-1) NaCl. Calculate the concentration of Na+ abd Cl- in mmol L-1 (assume solution density = 1.026 g mL-1)

2. If 0.1 mole of CaCl2 is dissolved in 10 L of water, calculate the resulting total solution concentrations Ca2+ and Cl- in mol kg-1 (ignore ion pairing, and assume solution density = 1 g mL-1).

3. If considering only carbon and oxygen, respiration can be represented by the schematic reaction equation:

CH2O + O2 --> CO2 + H2O

If respiration in an incubation bottle consumes 1.6 mg O2, calculate the mass of organic matter that has been respired (assumed organic matter has the composition CH2O) and the mass of CO2 that was liberated.

Explanation / Answer

1. A solution contains 35 parts per thousand (g kg-1) NaCl. Calculate the concentration of Na+ abd Cl- in mmol L-1 (assume solution density = 1.026 g mL-1)

Solution:- From the given data 35 g of solute are present in 1 kg of solution.

moles of NaCl = 35g * (1mol/58.44g) * (1000mmol/1mol) = 599 mmol

volume of solution = mass * density = 1 kg *(1000g/1kg) * (1ml/1.026g) * (1L/1000ml) = 0.975L

concentration of solution = 599mmol/0.975L = 614 mmol/L

NaCl -------> Na+ + Cl-

1 mmol of NaCl gives 1 mmol of Na+ and 1 mmol of Cl-. So, Na+ and Cl- concentrations will also be 614 mmol/L

2. If 0.1 mole of CaCl2 is dissolved in 10 L of water, calculate the resulting total solution concentrations Ca2+ and Cl- in mol kg-1 (ignore ion pairing, and assume solution density = 1 g mL-1).

Solution:- we have 0.1 mole of CaCl2.

Volume of solution is 10 L and density is 1 g/ml. So, we can calculate the mass of the solution.

10L * (1000ml/1L) * (1g/1ml) * (1kg/1000g) = 10kg

molality of CaCl2 = 0.1mole/10kg = 0.01 mole/kg

CaCl2 --------> Ca+2 + 2Cl-

1 mole of CaCl2 gives one mole of Ca+2 and two moles of Cl-

So, concentration of Ca+2 is 0.01 mole/kg and that of Cl- is 2*0.01 mole/kg = 0.02 mole/kg.

3. If considering only carbon and oxygen, respiration can be represented by the schematic reaction equation:

CH2O + O2 --> CO2 + H2O

If respiration in an incubation bottle consumes 1.6 mg O2, calculate the mass of organic matter that has been respired (assumed organic matter has the composition CH2O) and the mass of CO2 that was liberated.

Solution:- It is a stoichiometry problem. It's easy to solve the problem by using t-chart (dimensional analysis).

we will convert mg of O2 into g and then will calcukate its moles by diving its molar mass. Then we will use the mole ratio from balanced equation (1:1 mole ratio between CH2O and O2) and finally moles of CH2O will be converted into grams by multiplying its molar mass.

roadmap:- mg of O2 ----> g of O2 ---> moles of O2 ----> moles of CH2O -----> g of CH2O

1.6mg O2 * (1g/1000mg) * (1mole O2/32g O2) * (1 mole CH2O/1 mole O2) * (30g CH2O/1 mole CH2O) = 0.0015 g CH2O or 1.5 mg CH2O

We can make the set up for CO2 in the same way.

mg of O2 ----> g of O2 ---> moles of O2 ----> moles of CO2 ----> g of CO2

1.6 mg O2 * (1g/1000mg) * (1mole O2/32 g O2) * (1mole CO2/1 mole O2) * (44 g CO2/1 mole CO2) = 0.0022 g CO2 or 2.2 mg CO2.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.