Question

solve on paper
calculate the number of molecules in a)3 moles of sodium nitrate
b) 1.06 gram of sodium carbonate
c) 3.55 gram of clorine gas​

Answer 1

$\huge\mathfrak{\red{Answer}}$

a) 18.06× 10^23

b) 6.022 × 10^21

c) 59.6 × 10^21

Explanation:

To calculate the number of molecules we need to use the following method:

$\frac{given \: mass}{molecular \: mass} \times avogardo \: number$

The avogardo number is

$6.022 \times {10}^{23}$

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a) Given 3 moles of sodium carbonate

[To convert moles into molecular mass we multiply the given moles with the molecular mass]

3moles of sodium nitrate= 3× 85u=255g

Now,

No. of molecules=

$\frac{255}{85} \times 6.022 \times {10}^{23}$

$= 3 \times 0.622 \times {10}^{23} = 18.06 \times 10 {}^{23}$

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b) 1.06g of sodium carbonate

Given mass= 1.06g

molecular mass of sodium carbonate= 106u

avagardo number= 6.022×10^23

No. of molecules=

$\frac{given \: mass}{molecular \: mass} \times {6.022}^{} \times {10}^{23}$

$\frac{106}{106 \times 100} \times 6.022 \times {10}^{23}$

$= \frac{106}{106} \times 6.022 \times {10}^{21}$

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

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c) 3.55g of Chlorine gas

Given mass= 3.55g

molecular mass= 36u (approx.)

No. of molecules=

$\frac{355}{100 \times 36} \times 6.022 \times {10}^{23}$

$\frac{355}{36} \times 6.022 \times 10 {}^{21}$

$59.6 \times {10}^{21}$