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.

Answer 2

In the case of solid NaOH, it can be prepared by dissolving 40 gm in water and in the case 2 M NaOH it can be prepared by diluting 0.5 L of solution.

What is molarity?

Molarity of any substance is the concentration derived from the proportion of the moles and the volume of the solution in litres.

In the case of solid sodium hydroxide the mass needed is calculated as:

Given,

• Molar mass of NaOH = 40 g/mol
• Volume (V) of NaOH solution = 1.00 L
• Molarity (M) of the solution= 1.00 M
• Mass of NaOH = m

The formula for calculating molarity is:

$\rm Molarity = \dfrac{\rm mass}{\rm molar\; mass \times volume\; in\; L}$

Substituting values in the above equation:

$\begin{aligned}\rm 1.00\; M &= \dfrac{\rm m}{\rm 40\; g/mol \times 1.00 \;L}\\\\\rm m &= 1 \times 40\times 1\\\\&= 40 \;\rm g\end{alligned}$

Hence, for the solid sodium hydroxide, 40 gm of NaOH should be dissolved in some amount of water and should be made up to 1 L.

In the case of 2.00 M sodium hydroxide the volume needed is calculated as:

Given,

• Molarity of the NaOH solution $(\rm M_{1})$= 2.00 M
• The volume of NaOH solution $(\rm V_{1})$ = ?
• Molarity of solution after dilution $(\rm M_{2})$ = 1.00M
• Volume of solution after dilution $(\rm V_{2})$ = 1L

The dilution equation is given as:

$\rm M_{1}V_{1}= M_{2}V_{2}$

Substituting values in the equation:

$\begin{aligned}\rm V_{1} &= \dfrac{1.00 \rm\; M\times 1.00 \;\rm L}{2.00\;\rm M}\\\\&= 0.500 \rm \;L\end{aligned}$

Hence, a solution of 1.00 M concentration can be prepared by diluting 0.50 L of the 2 M NaOH solution in 1L of water.

Therefore, 40 gm and 0.500 L is the correct mass and volume needed.

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