Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsConcentration termEasy2 minPYQ_2020

ChemistryEasysingle choice

How many milli gram of iron $(F e^{2 +})$ are equal to 1 mL of $0.1055 N K_{2} C r_{2} O_{7}$ equivalent?

Answer:

Solution:

Milli eq. of Fe = Milli eq. of $K_{2} C r_{2} O_{7}$
$\frac{w}{56 / 1} \times 1000 = 1 \times 0.1055$
So $w = 5.9 \times 1 0^{- 3} g$
$= 5.9$ mg.

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Concentration TermsSubtopic:Concentration term

⏱ 2mℹ️ Source: PYQ_2020

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