JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Some Basic Concepts of Chemistry

250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ____

\times 1 0^{21}

(Nearest integer)

(N_{A} = 6.022 \times 1 0^{23})

Answer: 226

Solution

The amounts taken are

n_{NaOH} = 0.5 \times 0.250 = 0.125 mol, n_{HCl} = 1 \times 0.500 = 0.500 mol

Neutralisation runs mole for mole, so the alkali is consumed entirely and the acid surviving is

0.500 - 0.125 = 0.375 mol

. Multiplying by the Avogadro constant,

N = 0.375 \times 6.022 \times 1 0^{23} = 2.258 \times 1 0^{23} = 225.8 \times 1 0^{21}

To the nearest integer the number of unreacted HCl molecules is

226 \times 1 0^{21}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer