QJEEM04Ch607b: JEE Main 2004 Chemistry - Class 11 JEEM04Ch607b Solution: 6.02 × 10 20 molecules of urea are present in 100 ml of its…

Question

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6.02

\times

10²⁰ molecules of urea are present in 100 ml of its solution. The concentration of urea solution is
(Avogadro constant, N_A = 6.02

\times

10²³ mol^-1)

Accepted Answer

Moles (n) =

\frac{m o l e c u l e s}{N_{A}}

\Rightarrow

\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}}

\Rightarrow

0.001

Concentration (M) =

\frac{m o l e s}{v o l u m e}

Given volume of urea is 100 ml = 0.1 litre

[Note : While calculating Concentration volume should always be in litre.]

\Rightarrow

\frac{0.001}{0.1}

\Rightarrow

0.01 M

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