Question

32 g of oxygen contain same number of oxygen atoms as are present in x g of CaCO3, value of x is

Answer 1

➡ Given:

✏ 32g of oxygen contain same number os oxygen atoms as are present in x g of CaCO3.

➡ To Find:

✏ value of x

➡ Calculation:

$\leadsto \sf \: no.of \: o \: atoms = no.of \: o \: atoms \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: in \: O_2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: in \: CaCO_3 \\ \\ \leadsto \rm \: 2 \times \frac{weight}{Mw} \times N_A = 3 \times \frac{weight}{Mw} \times N_A$

• Weight of O2 = 32 g
• Mw of O2 = 32 g/mol
• Weight of CaCO3 = x g
• Mw of CaCO3 = 100 g/mol

$\leadsto \sf \: 2 \times \frac{ \cancel{32}}{ \cancel{32}} \times \cancel{N_A} = 3 \times \frac{x}{100} \times \cancel{N_A} \\ \\ \leadsto \rm \: 2 = \frac{3x}{100} \\ \\ \leadsto \rm \: x = \frac{200}{3 } \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{Weight \: of \: CaCO_3 = x = 66.67 \: g}}}}} \: \red{ \star}$