Question

Question #54463

Answer 1

`V_(NaCl)=0.34mL`
`V_(H_2O)=49.66mL`

Explanation:

The molarity of the solution is measured by: `C_M=(n_(NaCl))/(V_(sol))`

Therefore, the number of mole of sodium chloride is:

`n_(NaCl)=C_MxxV_(sol)=0.25(mol)/cancel(L)xx0.050cancel(L)=0.0125mol`

The mass of `NaCl` is calculated then by:
`m_(NaCl)=nxxMM=0.0125cancel(mol)xx58.45g/(cancel(mol))=0.731g`

The density of `NaCl` is: `d_(NaCl)=2.16" g/"mL`

The volume of `NaCl` is: `V_(NaCl)=m/d=(0.731cancel(g))/(2.16cancel(g)/(mL))=0.34mL`

The volume of water needed then:
`V_(H_2O)=V_("total")-V_(NaCl)=50.0mL-0.34mL=49.66mL`

There is no dilution factor since this a solution prepared from the solid solute and not from a stock solution.