Calculate the concentration of the Sulphamic acid solution.

Here are the calculations for the standardization of sodium hydroxide using sulphamic acid.

a) To find the concentration of the Sulphamic acid solution and the concentration of the NaOH solution in $\text{moldm}^{-3}$.

Step 1: Calculate the concentration of the Sulphamic acid solution.
From Table 1, the mass of Sulphamic acid used is $2.36 \text{ g}$.
The relative molecular mass of Sulphamic acid ($\text{NH}_2\text{SO}_3\text{H}$) is given as $97.1 \text{ g/mol}$.
The procedure states that the acid was dissolved and "made up to the mark", implying a standard solution was prepared. A common volume for such preparations is $250 \text{ cm}^3$.
Volume of Sulphamic acid solution = $250 \text{ cm}^3 = 0.250 \text{ dm}^3$.

Moles of Sulphamic acid = $\frac{\text{Mass}}{\text{Molar mass}}$

$$\text{Moles of Sulphamic acid} = \frac{2.36 \text{ g}}{97.1 \text{ g/mol}} \approx 0.0243048 \text{ mol}$$

Concentration of Sulphamic acid ($M_a$) = $\frac{\text{Moles}}{\text{Volume (dm}^3)}$

$$M_a = \frac{0.0243048 \text{ mol}}{0.250 \text{ dm}^3} \approx 0.0972192 \text{ mol/dm}^3$$

Rounding to three significant figures:

$$\boxed{\text{Concentration of Sulphamic acid} = 0.0972 \text{ mol/dm}^3}$$

Step 2: Calculate the concentration of the NaOH solution in $\text{moldm}^{-3}$.
The balanced chemical equation for the reaction is:

$$\text{NaOH(aq)} + \text{NH}_2\text{SO}_3\text{H(aq)} \longrightarrow \text{NH}_2\text{SO}_3\text{Na(aq)} + \text{H}_2\text{O(l)}$$

Sulphamic acid is a monobasic acid, so the mole ratio of $\text{NaOH}$ to $\text{NH}_2\text{SO}_3\text{H}$ is $1:1$.
Thus, $n_b = 1$ (for NaOH) and $n_a = 1$ (for Sulphamic acid).

From the titration data (Table 2):
Volume of Sulphamic acid ($V_a$) = $25 \text{ cm}^3$ (pipetted into the conical flask)
Mean titre volume of NaOH ($V_b$) = $25.2 \text{ cm}^3$ (average of the two accurate readings)

Using the titration formula:

$$\frac{M_a V_a}{n_a} = \frac{M_b V_b}{n_b}$$

Rearranging to solve for $M_b$:

$$M_b = \frac{M_a V_a n_b}{V_b n_a}$$

Substitute the values (using the unrounded $M_a$ for accuracy):

$$M_b = \frac{(0.0972192 \text{ mol/dm}^3) \times (25 \text{ cm}^3) \times 1}{(25.2 \text{ cm}^3) \times 1}$$ $$M_b = \frac{2.43048}{25.2} \text{ mol/dm}^3$$ $$M_b \approx 0.0964476 \text{ mol/dm}^3$$

Rounding to three significant figures:

$$\boxed{\text{Concentration of NaOH} = 0.0964 \text{ mol/dm}^3}$$

b) To calculate the concentration of the NaOH solution in $\text{gdm}^{-3}$.

Step 1: Determine the molar mass of $\text{NaOH}$.
Atomic mass of Na = $22.99 \text{ g/mol}$
Atomic mass of O = $16.00 \text{ g/mol}$
Atomic mass of H = $1.008 \text{ g/mol}$
Molar mass of $\text{NaOH}$ = $22.99 + 16.00 + 1.008 = 39.998 \text{ g/mol}$

Step 2: Calculate the concentration in $\text{gdm}^{-3}$.
Concentration ($\text{gdm}^{-3}$) = Molarity ($\text{mol/dm}^3$) $\times$ Molar mass ($\text{g/mol}$)
Using the unrounded molarity of NaOH:

$$\text{Concentration} = 0.0964476 \text{ mol/dm}^3 \times 39.998 \text{ g/mol}$$ $$\text{Concentration} \approx 3.8577 \text{ g/dm}^3$$

Rounding to three significant figures:

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