Question

A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would the student prepare this solution? If 2.00 MNaOH is avail- able, how would the student prepare the solution? To help insure three significant figures in the NaOH molarity, to how many sig- nificant figures should the volumes and mass be determined?

Answer 1

Explanation:

1)

$Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}$

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

$1.00 M=\frac{m}{40 g/mol\times 1.00 L}$

$m=1.00 M\times 40 g/mol\times 1.00 L = 40. g$

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

2)

$M_1V_1=M_2V_2$ (dilution equation)

Molarity of the NaOH solution = $M_1=2.00 M$

Volume of the solution = $V_1=?$

Molarity of the NaOH solution after dilution = $M_2=1.00 M$

Volume of NaOH solution after dilution= $V_2=1 L$

$M_1V_1=M_2V_2$

$V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L$

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.