Question

Solution Prep (a) The weight of MgCl2 required to create 1 kg of 1000 ppm Mg ion; (b)The weight of the 1000 ppm stock solution required to create a 15.0000 g of a 100 ppm sub-stock solution. (c) The weight of the 100 ppm solution required to create 15 g of each of the following: 1 ppm, 5 ppm, 10 ppm, 20 ppm, 30 ppm and 40 ppm Mg ion

Answer 1

a Ans.

1 ppm is nothing but 1mg/L

1 ppm is nothing but 1mg/L 1 ppm = 1mg/L Therefore, 1000ppm = 1000mg/L But, 1000mg = 1g Therefore, 1000mg = 1g/L So, the solution contain 1g of Mg ions to make 1000 ppm of Mg ions. Molar mass of MgCl_2 = 95.211g/mol That means, 24.305g of Mg is present in 95.211g of MgCl_2 Therefore, 1g of Mg will be present in mass of MgCl_2 given by = 95.211/24.305 times 1g = 3.917g of MgCl_2 Therefore, 1000 ppm of Mg ions = 3.917g of MgCl_2 in 1L(or 1kg) of water The 1000 ppm solution small amount of solute. So, 1L of the solution can be taken be equal to 1 1 kg = 1L 1000g = 1000 m L or, 1 g = 1 m L So, 15g of the second solution can be considered to be equal to 15mL. The amount of the 1000 ppm to be taken can be calculated using the formula below C_1 V_1 = C_2 V_2

Therefore,

But,

Therefore,

So, the solution should contain 1g of Mg ions to make 1000 ppm of Mg ions.

Molar mass of MgCl₂ = 95.211g/mol

That means, 24.305g of Mg is present in 95.211g of MgCl₂

Therefore, 1 g of Mg will be present in mass of MgCl₂ given by

of MgCl₂

Therefore, 1000 ppm of Mg ions = 3.917g of MgCl₂ in 1L (or 1kg) of water.

b Ans.

The 1000 ppm solution small amount of solute. So, 1L of the solution can be taken be equal to 1 kg

Or,

So, 15 g of the second solution can be considered to be equal to 15 mL.

The amount of the 1000 ppm to be taken can be calculated using the formula below

$C_{1}V_{1}=C_{2}V_{2}$

where C₁ and C₂ are the concentrations of the first and the second solution and V₁ and V₂ are the respective volumes.

$V_{1}=rac{C_{2}V_{2}}{C_{1}}$

Therefore, 1.5mL or 1.5g of the 1000 ppm stock must be taken in a 15mL volumetric flask and made upto the mark to make 15g of 100 ppm solution.

c Ans.

For all concentrations, you need to make 15g, which can be taken to be 15mL.

Use the same formula to calculate for each concentration.

$C_{1}V_{1}=C_{2}V_{2}$

$V_{1}=rac{C_{2}V_{2}}{C_{1}}$

From here, C₁ and V₁ are the concentration and volume of 100 ppm solution.

C₂ and V₂ are the concentrations and volume of the new solution to be made.

To make 15g of 1ppm solution.

$V_{1}=rac{C_{2}V_{2}}{C_{1}}$

Therefore, 0.15g or 15 mL of 100 ppm solution must be taken in a 15 mL volumetric flask to make 15g of 1 ppm solution.

5 ppm

$V_{1}=rac{C_{2}V_{2}}{C_{1}}$

Therefore, 0.75 mL or 0.75g of 100 ppm solution must be taken in 15mL volumetric flask to make 15g of 5 ppm solution.

Similarly, it can be calculated for 10ppm, 20ppm, 30ppm and 40ppm. It will be found to 1.5g, 3.0g, 4.5g and 6.0g respectively.