Calculate the number of moles- molecules- present in 1-1 - 10-4 kg of CO2-

1.1× 10 ^ 4 kg of CO _ 2 (M.W of CO _ 2 =44 g) ∴ No. of moles = 1.1× 10 ^ 4 × 10 ^ 3 g 44 =2.5× 10 ^ 3 ∴ No. of molecules ...

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Standard X

Chemistry

Mole Concept and Avogadro's Number

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Question

Calculate the number of moles, molecules, present in $1.1 \times 10^{- 4} k g$ of $C O_{2}$ .

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Solution

$1.1 \times 10^{- 4}$ kg of ${C O}_{2}$ (M.W of ${C O}_{2} = 44$ g)
$∴$ No. of moles $= \frac{1.1 \times 10^{- 4} \times 10^{3} g}{44} = 2.5 \times 10^{- 3}$
$∴$ No. of molecules $=$ No. of moles $\times N_{A}$
$= 2.5 \times 10^{- 3} \times 6.022 \times 10^{23} = 15.055 \times 10^{20}$

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Calculate the number of moles, molecules, present in 1.1×10−4kg of CO2.

1.1×10−4kg of CO2 (M.W of CO2=44g)∴ No. of moles=1.1×10−4×103g44=2.5×10−3∴ No. of molecules=No. of moles ×NA=2.5×10−3×6.022×1023=15.055×1020

Calculate the number of moles, molecules, present in $1.1 \times 10^{- 4} k g$ of $C O_{2}$ .

$1.1 \times 10^{- 4}$ kg of ${C O}_{2}$ (M.W of ${C O}_{2} = 44$ g)
$∴$ No. of moles $= \frac{1.1 \times 10^{- 4} \times 10^{3} g}{44} = 2.5 \times 10^{- 3}$
$∴$ No. of molecules $=$ No. of moles $\times N_{A}$
$= 2.5 \times 10^{- 3} \times 6.022 \times 10^{23} = 15.055 \times 10^{20}$