Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsMiscellaneous/MixedMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

A sample of urine containing 0.3 g urea was treated with excess 0.2 M nitrous acid. $NH_{2} CONH_{2} + 2 HNO_{2} \to CO_{2} + 2 N_{2} + 3 H_{2} O$ The gases produced are passed through aq. KOH and final volume is measured. ( $R = 0.082 L -$ atm $/ K - mol$ )

Volume of final gas (or gaseous mixture) at 27°C and 1 atm is $____$ (in mL)

Answer:

246.00

Solution:

Moles of urea:

$\frac{0.3}{60} = 0.005 m o l$

From the reaction:

$1 m o l u re a \to 2 m o l N_{2}$

So, the moles of N₂ are:

$N_{2} = 0.005 \times 2 = 0.01 m o l$

After passing through KOH, CO₂ is absorbed, so only N₂ remains.

Using the ideal gas law, $P V = n RT$ :

$V = 0.01 \times 0.082 \times 300 = 0.246 L$

$V = 246 m L$

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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