Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2439 Upvotes
Verified Answer
The correct answer is:
250 mg
$\begin{aligned}
& \mathrm{M}=\frac{\mathrm{W} \times 1000}{\left.\mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)} \\
& \mathrm{W}=\frac{\left.\mathrm{M} \times \mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)}{1000}=\frac{0.75 \times 36.5 \times 25}{1000} \\
&=0.684 \mathrm{~g} \text { (Mass of } \mathrm{HCl}) \\
& \underset{36.5 \mathrm{~g}}{\mathrm{HCl}}+\underset{40 \mathrm{~g}}{\mathrm{NaOH}} \longrightarrow \mathrm{HCl}+\mathrm{NaOH}
\end{aligned}$
36.5 g HCl reacts with $\mathrm{NaOH}=40 \mathrm{~g}$
0.684 g HCl reacts with $\mathrm{NaOH}=\frac{40}{36.5} \times 0.684 \simeq 0.750 \mathrm{~g}$
Amount of NaOH left $=1 \mathrm{~g}-0.750 \mathrm{~g}=0.250 \mathrm{~g}=250 \mathrm{mg}$
& \mathrm{M}=\frac{\mathrm{W} \times 1000}{\left.\mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)} \\
& \mathrm{W}=\frac{\left.\mathrm{M} \times \mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)}{1000}=\frac{0.75 \times 36.5 \times 25}{1000} \\
&=0.684 \mathrm{~g} \text { (Mass of } \mathrm{HCl}) \\
& \underset{36.5 \mathrm{~g}}{\mathrm{HCl}}+\underset{40 \mathrm{~g}}{\mathrm{NaOH}} \longrightarrow \mathrm{HCl}+\mathrm{NaOH}
\end{aligned}$
36.5 g HCl reacts with $\mathrm{NaOH}=40 \mathrm{~g}$
0.684 g HCl reacts with $\mathrm{NaOH}=\frac{40}{36.5} \times 0.684 \simeq 0.750 \mathrm{~g}$
Amount of NaOH left $=1 \mathrm{~g}-0.750 \mathrm{~g}=0.250 \mathrm{~g}=250 \mathrm{mg}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.