Question

In the making stock solutions from solids problem, make a 100 ml solution that is 2.5 M of the Cl- ion using NaCl and make a 100 mL solution that is 0.5 M of the Cl- ion using MgCl2. Show your calculations and give details about the equipment used, which solutions you mixed together, and amounts of each.

Answer 1

Step-by-step explanation:

`Molarity=(Moles)/(Volume (L))`

1) Molarity of the chloride solution = 2.5 M

Volume of the chloride solution = 100 mL = 0.100 L ( 1mL = 0.001 L)

Moles of chloride ions = n

`2.5 M=(n)/(0.100 L)`

`n=2.5M* 0.100 L=0.25 mol`

`NaCl(aq)\rightarrow Na^+(aq)+Cl^-(aq)`

1 mole of chloride ions are obtained from 1 mole NaCl,then 0.25 mol of chloride ions will be obtained from :

`1* 0.25 mol=0.25 mol` of NaCl

Mass of 0.25 moles of NaCl :

0.25 mol × 58.5 g/mol = 14.62 g

In a dry 100 mL volumetruc flask add weighed 14.62 grams of NaCl and small amount of water to dissolve it completely. After this add the water upto mark of 100 mL to form chloride solution of 0.25 M.

2) Molarity of the chloride solution = 0.5 M

Volume of the chloride solution = 100 mL = 0.100 L ( 1mL = 0.001 L)

Moles of chloride ions = n

`0.5 M=(n)/(0.100 L)`

`n=0.5M* 0.100 L=0.05 mol`

`MgCl_2(aq)\rightarrow Mg^(2+)(aq)+2Cl^-(aq)`

2 mole of chloride ions are obtained from 1 mole
`MgCl_2`,then 0.05 mol of chloride ions will be obtained from :

`(1)/(2)* 0.05 mol=0.025 mol` of
`MgCl_2`

Mass of 0.025 moles of
`MgCl_2` :

0.025 mol × 95 g/mol = 2.375 g

In a dry 100 mL volumetruc flask add weighed 2.375 grams of
`MgCl_2` and small amount of water to dissolve it completely. After this add the water upto mark of 100 mL to form chloride solution of 0.5 M.