1.

Calculate the number of molecules in 1.195 g of chloroform. (CHCl3).(At. mass, H = 1u, C = 12u, Cl = 35.5 u)

Answer»

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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Warm REGARDS

@Laughterqueen

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