Calculation of mass from Avogadro's number — lesson. Science State Board, Class 10.

PDF chapter test TRY NOW

1.	Calculate the mass of \(1.51\times10^{23}\) molecules of water.

Data:

Given molecules of \(H_2O=1.51\times10^{23}\)

Molecular mass of \(H_2O=18\)

Avogadro's number \(=6.023\times10^{23}\)

Solution:

\begin{array}{l} Number of moles & = & \frac{Number of molecules of water}{Avogadro number} \\ = & \frac{1.51 \times 10^{23}}{6.023 \times 10^{23}} \\ = & \frac{1}{4} \\ = & 0.25 mole \end{array}

\begin{array}{l} Number of moles & = & \frac{Mass}{Molecular mass} \\ 0.25 & = & \frac{Mass}{18} \\ Mass & = & 0.25 \times 18 \\ Mass & = & 4.5 g \end{array}

Hence, the mass of \(1.51\times10^{23}\) molecules of water \(=4.5\) \(g\)

2.	Calculate the mass of \(5\times10^{23}\) molecules of glucose.

Data:

Given molecules of glucose \(=1.51\times10^{23}\)

Molecular mass of glucose \(=180\)

Avogadro's number \(=6.023\times10^{23}\)

Formula:

\begin{array}{l} Mass of glucose & = & \frac{Molecular mass \times Number of particles}{Avogadro number} \\ = & \frac{180 \times 5 \times 10^{23}}{6.023 {\times 10}^{23}} \\ = & 149.43 g \end{array}

Hence, the mass of \(5\times10^{23}\) molecules of glucose \(=149.43\) \(g\)

Did you find an error?

5. Calculation of mass from Avogadro's number