Question

calculate the ratio of molecules present in 44 g of carbon-dioxide and 16 g of SO2​

Answer 1

★Solution :-

♠ Carbon dioxide :-

• Mass of CO 2 = 44 g
• Molar mass of CO 2 = 44 g

Calculation :-

• Molar mass of CO 2 = C + 2(O) = 12+ 32 = 44 g

we know that ,

$\bigstar \boxed{\mathtt{ moles= \dfrac{mass}{molar mass} = \dfrac{Molecules }{6.023 \times {10}^{23} }}}$

hence ,

$\implies \mathtt{ \dfrac{mass_{CO_2}}{molar \: mass_{CO_2}} = \dfrac{Molecules \: of \: CO_2 }{6.023 \times {10}^{23} }}$

$\implies \mathtt{ Molecules \: of \: CO_2 =\dfrac{mass_{CO_2}}{molar \: mass_{CO_2}} \times 6.023 \times {10}^{23} }$

$\implies \mathtt{ Molecules \: of \: CO_2 =\dfrac{44}{44} \times 6.023 \times {10}^{23} }$

$\implies \mathtt{ Molecules \: of \: CO_2 = 6.023 \times {10}^{23} }$

♠ Sulphur dioxide :-

• Mass of SO 2 = 16 g
• Molar mass of SO 2 = 64 g

Calculation :-

Molar mass of SO 2 = S + 2(O) = 32+ 32 = 64 g

we know that ,

$\implies \mathtt{ \dfrac{mass_{SO_2}}{molar \: mass_{SO_2}} = \dfrac{Molecules \: of \: SO_2 }{6.023 \times {10}^{23} }}$

$\implies \mathtt{ Molecules \: of \: SO_2 =\dfrac{mass_{SO_2}}{molar \: mass_{SO_2}} \times 6.023 \times {10}^{23} }$

$\implies \mathtt{ Molecules \: of \: SO_2 =\dfrac{16}{64} \times 6.023 \times {10}^{23} }$

$\implies \mathtt{ Molecules \: of \: SO_2 =\dfrac{1}{4} \times 6.023 \times {10}^{23} }$

$\implies \mathtt{ Molecules \: of \: SO_2 =1.505\times {10}^{23} }$

♠ Ratio of the molecules of CO 2 and SO 2 :-

$\implies\mathtt{\dfrac{Molecules \: of \: CO_2}{Molecule s \: of \: SO_2} = \dfrac{6.023 \times {10}^{23} }{1.505 \times {10}^{23} } }$

$\implies \boxed{\mathtt{\dfrac{Molecules \: of \: CO_2}{Molecule s \: of \: SO_2} =4}}$