Question

Calculate the number of molecules in 1.195 g of chloroform. (CHCl3).

(At. mass, H = 1u, C = 12u, Cl = 35.5 u)

Answer 1

Heya mate,Here is ur answer

Molar mass of CHCl3 = Mass of C+ Mass of H + 3× Mass of Cl

Molar Mass of CHCl3 = 12+1+3×35.5

Molar mass of CHCl3 = 12+1+106.5

Molar mass of CHCl3 = 119.5 g.

We know that each mole of any compound has fixed number of 6.022×10^23 molecules. Here 6.022×10^23 is known as " Avagadro Number " .

So, 119.5 g of CHCl3 has = 6.022 × 10^23 molecules

1 gm of CHCl3 has = 6.022×10^3 / 119.5 molecules.

1.195 g of CHCl3 has =
$\frac{6.022 \times 10 {}^{23} }{119.5} \times 1.195$

$= \frac{6.022 \times 10 {}^{23} \times 10 }{1195 \times 1000} \times 1195$


$= \frac{6.022 \times 10 {}^{24} }{10 {}^{3} }$


$= 6.022 \times 10 {}^{21}$

So, 1.195 g of CHCl3 has 6.022 ×10^21 molecules.

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@Laughterqueen

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Answer 2

Heyaa☺

➡Refer to the above attachment.

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Hope you understand..✌
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