SSC
CHEMISTRY
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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cm3 of 0.3 mol dm-3 sodium hydroxide neutralises 15 cm3 of hydrochloric acid solution. Determine the concentration of the hydrochloric acid solution.
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0.25 mol dm-3
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0.3 mol dm-3
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0.5 mol dm-3
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1.0 mol dm-3
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Explanation:
Detailed explanation-1: -In this case, we need to find the concentration in grams per deciliter. To do this, divide 0.10 mol/dm3 by 1 liter. Therefore, the concentration of the sodium hydroxide solution is 10 g/dm3.
Detailed explanation-2: -What Happened in the Titration: In the reaction between NaOH and HCl, 1 mole of NaOH reacts with 1 mole of HCl. 1 NaOH + 1 HCl → 1 NaCl + 1 H2O (Mole ratio of NaOH to HCl is 1:1) • The concentration of the NaOH was 0.1 M, or 0.1 moles/liter.
Detailed explanation-3: -Total acidity (as Hcl) percent by mass = (VXNX3. 646)/M. V = Volume in ml of standard Sodium Hydroxide solution Used in titration N = Normality of standard Sodium Hydroxide solution M = Mass in grams of the sample taken for the test.
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