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Match the following
List -I List-II
1) Total number of atoms in 64 gm $S O_{2}$ is A) $6.023 \times 10^{20}$
2) no. of oxalic acid molecules in 100 ml of 0.01 M oxalic acid B) $0.0196$
3) 5.85gm $N a C l$ are dissolved in 90 gm of water. The mole fraction is C) $1.66 \times 10^{- 24}$
4) Mass of 1 atom of hydrogen is D) $3 N_{o}$
Choose the correct answer from the options given below

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$1 - d; 2 - b; 3 - c; 4 - a$

$1 - d; 2 - b; 3 - c; 4 - d$

$1 - d; 2 - a; 3 - b; 4 - c$

$1 - d; 2 - a; 3 - c; 4 - b$

answer is C.

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Detailed Solution

1) Total no. of atoms $= a t o m a c i t y \times n \times N o = 3 \times \frac{64}{64} \times N o$

$2) M = \frac{n}{v} = n / \frac{100}{100} \Rightarrow n = 0.001$

Molecules $= n \times N o = 6.023 \times 10^{23} \times 0.001 = 6.023 \times 10^{20}$

$3) x_{N a c l} = \frac{\frac{5.85}{58.5}}{\frac{5.85}{58.5} + \frac{90}{18}} \Rightarrow 0.0196$

$4) 6.023 \times 10^{23} - 1 g m$

$1 a t o m = 1 / 6.023 \times 10^{23} = 1.66 \times 10^{- 24}$

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