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A flask contains 3.0xx10^(16) atoms of H...

A flask contains `3.0xx10^(16)` atoms of He. From This `6.6xx10^(15)` atoms of He are removed. The flask now contains atoms of He

A

`3.6xx10^(16)`

B

`2.3xx10^(16)`

C

`+3.6xx10^(15)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`3.0xx10^(16)-6.6xx10^(15)`
`=30xx10^(15)-6.6xx10^(15)=(30-6.6)xx10^(15)`
`=23.4xx10^(15)=2.34xx10^(16)=2.3xx10^(16)`
(After rounding off the last digit)
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