Question 1.28
NCERT Class XI Chemistry
Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g)
Video Explanation
(detailed solution given after this video)
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Solution in Detail
$\displaystyle \text{no. of moles } = \frac{\text{w (mass)}}{\text{M (molar mass)}}$
$\displaystyle \therefore \text{no. of atoms } = \frac{\text{w}}{\text{M}} \times 6 \times 10^{23}$
but all masses are given same w = 1 gram.
$\displaystyle \therefore \text{atoms} \propto \frac{1}{M}$
$\displaystyle \therefore$ the one with lowest molar mass will have the most number of atoms
Molar masses are Au = 197, Na = 23, Li = 7, Cl2 = 35.5 x 2 = 71 (all in (g/mol) )
Hence lithium, Li, is the answer!
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