Question

calculate the no. of Molecules present in 9.8g of sulphuric acid​

Answer 1

Number of molecules(N):

$\boxed{ \bf N = n\times N_{A}}$

where n = no. of moles , N_A = avagadro constant = 6.023 × 10²³

Solution

$\rm Sulphuric \: acid = H_2SO_4$

• molar mass = 98

$\to\rm moles = \dfrac{mass}{molar \: mass}$

$\to\rm n = \dfrac{m}{M} = \dfrac{9.8}{98}$

$\to\bf n = 0.1$

-----

$\to\rm N = 0.1\times 6.023 \times 10^{23}$

$\to\bf N = 6.023 \times 10^{22}$

★ so, number of molecules present in 9.8 g of Sulphuric acid is 6.023 × 10²².

Answer 2

Answer:

• Given mass of sulphuric acid = 9.8 g

First let's calculate the molar mass of Sulphuric acid:

$\footnotesize\longrightarrow\:\sf M_{(H_2SO_4)} = 2 \times 1 + 32 + 16 \times 4$

$\footnotesize\longrightarrow\:\sf M_{(H_2SO_4)} = 2 + 32 + 64$

$\footnotesize\longrightarrow\: \underline{ \underline{\sf M_{(H_2SO_4)} = 98 \: \frac{g}{mol}}}$

Now, let's find the no. of moles of Sulphuric acid:

$\footnotesize\longrightarrow\:\sf n_{(H_2SO_4)} = \frac{Mass}{Molar \:Mass }$

$\footnotesize\longrightarrow\:\sf n_{(H_2SO_4)} = \frac{9.8}{98 }$

$\footnotesize\longrightarrow\: \underline{ \underline{\sf n_{(H_2SO_4)} = 0.1 \: mol }}$

Now, we can calculate the no. of molecules in Sulphuric acid:

$\footnotesize\longrightarrow\:\sf n_{(H_2SO_4)} = \frac{Molecules}{Avogadro's \: Number }$

$\footnotesize\longrightarrow\:\sf 0.1 = \frac{Molecules}{6.022 \times {10}^{23} }$

$\footnotesize\longrightarrow\:\sf Molecules = 0.1 \times 6.022 \times {10}^{23}$

$\footnotesize\longrightarrow\:\sf Molecules = 0.6022 \times {10}^{23}$

$\footnotesize\longrightarrow\:\sf Molecules = 6.022 \times {10}^{23} \times {10}^{ - 1}$

$\footnotesize\longrightarrow\: \underline{ \orange{ \boxed{ \gray{\bf Molecules = 6.022 \times {10}^{22} }}} }$