2.8 10⁻³ ~mol of CO ₂ is left after removing 10²¹ molecules from its ' x ' mg sample. The mass of CO ₂ taken initially is Given: N _ A =6.02 10²³ ~mol ⁻¹

Step-by-step solution

Moles of removed CO ₂= 10²¹ 6.02 10²³ ~mol =1.66 10⁻³ ~mol mole of CO ₂ left =2.8 10⁻³ moles total moles of CO ₂ taken initially =(2.8+1.66) 10⁻³ ~mol mass of CO ₂ taken initially aligned & =4.46 10⁻³ 44 & =196.24 10⁻³ ~g & =196.24 mg aligned