Question

The number of molecules of CaCO3, present in
100 mL of 0.01M Caco3, solution is
(1) 6.022 x 10^22
(2) 6.022 x 10^20
(3) 6.022 x 10^26
(4) 6.022 10^18​

Answer 1

Explanation:

m=mole/volume in liter

0.01=mole/0.1

mole=0.01×0.1

mole =0.001

mole=10^-3

no. of molecule =mole ×6.022×10^23

no. of molecule =10^-3 ×6.022 ×10^23

no of molecule =6.022×10^20

hence option 2 is correct

hope u got ur answer

Answer 2

Answer:

here is the answer :

Explanation:

m = mole / volume in liter

0.01 = mole/0.1

mole = 0.01×0.1

mole = 0.001

mole = 10^-3

no. of molecules = mole × 6.022 × 10^23

no. of molecules = 10^-3 × 6.022 × 10^23

no. of molecules = 6.022 × 10^20

so, here option 2 is correct.

the answer will be : 6.022 × 10^20.

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