1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

The correct option is 250 mg.

Step-by-step explanation:

To find the mass of sodium hydroxide ($\mathrm{NaOH}$) left unreacted, we can follow these steps:

1. Write the balanced chemical equation for the reaction:
The neutralization reaction between sodium hydroxide ($\mathrm{NaOH}$) and hydrochloric acid ($\mathrm{HCl}$) is given by:
$\mathrm{NaOH}+\mathrm{HCl}\to \mathrm{NaCl}+H_{2}O$
From the equation, 1 mole of $\mathrm{NaOH}$ reacts completely with 1 mole of $\mathrm{HCl}$.

2. Calculate the initial number of moles of $\mathrm{NaOH}$:
The molar mass of $\mathrm{NaOH}$ is calculated as:
$23\text{(Na)}+16\text{(O)}+1\text{(H)}=40\text{g/mol}$
Given mass of $\mathrm{NaOH}$ = 1 g.
The initial moles of $\mathrm{NaOH}$ are:
$\text{Initial Moles of NaOH}=\frac{1\text{g}}{40\text{g/mol}}=0.025\text{moles}$
Converting this to millimoles (mmol):
$0.025\times 1000=25\text{mmol}$

3. Calculate the number of moles of $\mathrm{HCl}$ added:
Volume of $\mathrm{HCl}$ solution = 25 mL
Molarity of $\mathrm{HCl}$ solution = 0.75 M
The moles of $\mathrm{HCl}$ are:
$\text{Moles of HCl}=\text{Molarity}\times \text{Volume in L}$
$\text{Moles of HCl}=0.75\text{mol/L}\times 0.025\text{L}=0.01875\text{moles}$
Converting this to millimoles (mmol):
$0.01875\times 1000=18.75\text{mmol}$

4. Calculate the unreacted amount of $\mathrm{NaOH}$:
Since the reaction ratio is 1:1, 18.75 mmol of $\mathrm{HCl}$ will neutralize exactly 18.75 mmol of $\mathrm{NaOH}$.
Therefore, the amount of $\mathrm{NaOH}$ left unreacted is:

5. Convert the unreacted moles back to mass in milligrams:
The mass of unreacted $\mathrm{NaOH}$ is:
$\text{Mass of unreacted NaOH}=6.25\text{mmol}\times 40\text{g/mol}=250\text{mg}$