A chemist must prepare 975. mL of 160, mM aqueous sodium carbonate (Na_2CO_3) working solution. He'll do this by pouring out some 0.174 mol/L aqueous sodium carbonate stock…

Question

A chemist must prepare 975. mL of 160, mM aqueous sodium carbonate (
`Na_2CO_3`) working solution. He'll do this by pouring out some 0.174 mol/L aqueous sodium carbonate stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the sodium carbonate stock solution that the chemist should pure out. Round your answer to 3 significant digits.

Answer 1

897 mL

Step-by-step explanation:

This is a problem of dilution. According to the dilution principle, the number of mole before dilution must be equal to the number of mole after dilution. Mathematically:

Molarity x volume (before dilution) = molarity x volume (after dilution)

Before dilution:

Molarity = 0.174 mol/L

Volume = ?

After dilution:

Molarity = 160 mM = 160/1000 = 0.160 M

Volume = 975 mL

Substitute into the dilution equation:

0.174 x volume of sodium carbonate stock = 0.160 x 975

Volume of stock = 0.160 x 975/0.174

= 896.552

To 3 significant digits = 897 mL

The chemist will pour out 897 mL stock solution of sodium carbonate.