JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
\(250\, \mathrm{~mL}\) of \(0.5\, \mathrm{M}\, \mathrm{NaOH}\) was added to \(500\, \mathrm{~mL}\) of \(1\, \mathrm{M}\, \mathrm{HCl}\) The number of unreacted \(\mathrm{HCl}\) molecules in the solution after complete reaction is \(......\,\times 10^{21}\). (Nearest integer) \(\left(\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23}\right)\)
- A \(226\)
- B \(235\)
- C \(462\)
- D \(521\)
Answer & Solution
Correct Answer
(A) \(226\)
Step-by-step Solution
Detailed explanation
We known that no. of moles \(=\text { Vlitre } \times \text { Molarity and No. of millimoles }\) \(=V_{m l} \times \text { Molarity }\) \(\text { so millimoles of } \mathrm{NaOH}=250 \times 0.5=125\) \(\text { Millimoles of } \mathrm{HCl}=500 \times 1=500\)…
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