How many grams of urea on heating yield $10^{22}$ molecules of biuret by the reaction given below?
$2 C O (N H_{2})_{2} ⟶ H_{2} N - C O - N H - C O - N H_{2} + N H_{3}$ ?

1.495

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0.995

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1.99

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1.753

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Solution

The correct option is C $1.99$
$10^{22}$ molecules of biuret corresponds to $\frac{10^{22}}{6.023 \times 10^{23}} = 0.0166$ moles of biuret.
They will be obtained by heating $2 \times 0.0166 = 0.033$ moles of urea.
This corresponds to $60 \times 0.033 = 1.99$ g of urea.

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