Question

A solution of NaOH is 4 g L–1. What volume of HCl gas at STP will neutralize, 50 ml of the alkali solution?
(1) 224 ml (2) 112 ml (3) 11200 ml (4) 22.4 ml

Answer 1

Answer:

Molar mass of NaOH = 40 g/mol

Mass of NaOH involved in 50ml alkali solution is 4/1000 × 50 = 0.2g

Moles of NaOH in reaction = 0.2/40 = 0.005

Now we know that 1 mol of HCl completely neutralize 1 mol NaOH

So, 0.005 moles HCl will neutralize 0.005 moles of NaOH.

At STP, volume of 1 mol HCl is 22.4L

So, volume of 0.005 mol HCl is 22.4×0.005 = 0.112L = 112ml.

Option (2) is correct.

Answer 2

Answer:

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