Physics, asked by chiragmcmahon, 9 months ago

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

Answers

Answered by Anonymous
3

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}

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