Question

How many moles of NaCl is needed to prepare 67.3g of 21% solution

Answer 1

The total number of moles for making 21%
`NaCl` solution is :
`0.1997mol`.

Step-by-step explanation:

Given :

• 21% solution
• Weight =
  `67.3g`
• Weight of
  `NaCl=23+35.5=58.5`

If a solution is '
`x`'% , it means :
`'x' g` of the solute is present in
`100ml` of solution.

Here the solvent is water.

∴

21% Solution :

`(21gNaCl)/(100mlH_(2)O)`

Since the volume of water doesn't change much as the salt dissolves or ionizes ,

we can assume that we have used
`100g` of
`H_(2)O` has been used for making this solution.

• thus , EVERY
  `121g` of the solution , contains
  `21g` of
  `NaCl`.

thus
`67.3g` will have :

`(67.3)/(121) *21//=11.68g`

Thus , number of moles :

`(total.weight)/(weight.per.mole)//=(11.68)/(58.5)//=0.1997mol`

Answer :
`0.1997mol`

Answer 2

The required number of moles of NaCl is 0.2416 moles

Step-by-step explanation:

The first thing to note here is the given weight (67.3g) of the total solution. The 21% also tells us the weight condition of NaCl in the entire solution, bearing in mind that the remaining 79% (gotten from 100% - 21%) will be the solvent in the solution.

So, we find the weight of NaCl in the solution (Please see attached)

Then we calculate the Molar mass of NaCl (Please see attached)

Finally, the number of moles can be obtained.

Please find attached a detailed solution, clearly showing the steps.