Question

6.3145 g of NaCl was dissolved in enough water to make 100.0 mL of solution. If you pipet a 25.00 mL portion of this solution to an evaporating dish, and evaporate off the water to dryness, what mass of NaCl would you expect to collect?

Answer 1

Following is the - complete Answer -&- Explanation: for the given Question, in.....typed format....

$\Rightarrow$Answer:

Mass of NaCl, we should expect to collect =  1.578 g ( grams ), from the dried off evaporating disk.

$\Rightarrow$Explanation:

Following is the complete Explanation, for the above Answer...

• Given:

1. Mass of sodium chloride ( NaCl , solute ) = 6.3145 g ( grams )
2. Volume of the solution: V_sol = 100.0 mL
3. Assuming, that in this case, the desired mass, of NaCl, is dissolved in 100.0 mL Water, to produce a completely homogeneous solution.

• ​​​​​​​Step - 1:

​​​​​​​From the given information, we have come to know the following: i.e.

$\Rightarrow$ 100.0 mL of the solution, contains 6.3145 g ( grams ) of the solute, i.e. NaCl

$\Rightarrow$25.0 mL of the solution, would contain: [ ( 6.3145 g ) / ( 100.0 mL ) ] x ( 25.0 mL ) = 1.578 g ( grams ), of the solute, i.e. NaCl ...

• Step - 2:

​​​​​​​Therefore, according to the above discussion, we can surely say that, if we pipette out , a portion: i.e. 25.0 mL , of the given homogeneous aqueous solution of sodium hydroxide ( NaOH ) , and place it on an evaporating disk, and evaporate off, all the water ( solvent ), from the evaporating disk, we would obtain the dry mass, of the dry solute ( i.e. NaCl ), dissolved in 25.0 mL , of the given aqueous solution....

Therefore, we can say -- we would be able to collect:  1.578 g ( grams ) of dry solid NaCl, after drying out the evaporating disk, containing 25.0 mL of the prepared solution... as discussed above...

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