In an experiment to determine the molar concentration of a standard solution in the laboratory, different volumes of standard sodium hydroxide solution were dried to constant masses. The dried masses(M/g) of the solute sodium hydroxide and the corresponding volumes(V/cm cube) of the standard solution is as follows : Mass of NaOH=8,30,50,70,80 respectively. The volume of solute=40,150,250,350,400,respectively. a. Plot a graph of mass(M/g) on the vertical axis and volume(V/cm cube) on horizontal axis. b.( I) determine the slope of the graph. (Ii) state the significant of the slope. c. From the graph, determine the mass of the solute in 200cm cube of solution. d. Calculate the molar concentration of the sodium hydroxide in the standard solution. (Na=23,O=16,H=1) c.State two precautions to be taken when preparing the standard solution.

Draw a horizontal axis for Volume (V/cm$^3$) ranging from 0 to at least 400 cm$^3$.
Draw a vertical axis for Mass (M/g) ranging from 0 to at least 80 g.

Plot the following points:

• $(40 \text{ cm}^3, 8 \text{ g})$
• $(150 \text{ cm}^3, 30 \text{ g})$
• $(250 \text{ cm}^3, 50 \text{ g})$
• $(350 \text{ cm}^3, 70 \text{ g})$
• $(400 \text{ cm}^3, 80 \text{ g})$

Draw a straight line that passes through all the plotted points and extends to the origin $(0,0)$.

We can use any two points from the given data, for example, $(V_1, M_1) = (40 \text{ cm}^3, 8 \text{ g})$ and $(V_2, M_2) = (400 \text{ cm}^3, 80 \text{ g})$.

The slope $m$ is given by the formula:

Substitute the chosen points:

The slope of the graph is $\boxed{\text{0.2 g/cm}^3}$.

(II) State the significance of the slope.
The slope of the mass versus volume graph represents the mass concentration (or density) of the sodium hydroxide solution. It indicates the mass of NaOH solute present per unit volume of the solution.

Since the graph is a straight line passing through the origin, its equation is $M = m \times V$, where $m$ is the slope.
We found the slope $m = 0.2 \text{ g/cm}^3$.

The mass of the solute in $200 \text{ cm}^3$ of solution is $\boxed{\text{40 g}}$.

Molar mass of NaOH = Atomic mass of Na + Atomic mass of O + Atomic mass of H
Molar mass of NaOH = $23 \text{ g/mol} + 16 \text{ g/mol} + 1 \text{ g/mol}$
Molar mass of NaOH = $40 \text{ g/mol}$

Mass concentration = $0.2 \text{ g/cm}^3$
Since $1 \text{ L} = 1000 \text{ cm}^3$:
Mass concentration in g/L = $0.2 \text{ g/cm}^3 \times 1000 \text{ cm}^3/\text{L}$
Mass concentration in g/L = $200 \text{ g/L}$

Molar concentration = $\frac{\text{Mass concentration in g/L}}{\text{Molar mass in g/mol}}$
Molar concentration = $\frac{200 \text{ g/L}}{40 \text{ g/mol}}$
Molar concentration = $5 \text{ mol/L}$
The molar concentration of the sodium hydroxide in the standard solution is $\boxed{\text{5 mol/L}}$.