The number of gram -molecules of oxygen in 6.02 \times1024 carbon monoxide molecules is :

Foundation

Chemistry Foundation

Mole Concept and Calculation

Question

The number of gram -molecules of oxygen in 6.02 $\times$ 10²⁴ carbon monoxide molecules is :

5.0 gram molecules

0.5 gram molecules

10.0 gram molecules

1.0 gram molecules

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

∵ 6.023 × 10²³ molecules of CO = 1 g molecule.
∴ 6.023 × 10²⁴ molecules of CO will be
$= \frac{1}{6 .023 \times 10^{23}} \times 6 .023 \times 10^{24}$
= 10g molecules