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  • 3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are (a) 6.68\times 10^{23} (b) 6.09\times 10^{22} (c) 6.022\times 10^{23}

3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are

(a) 6.68\times 10^{23}
(b) 6.09\times 10^{22}
(c) 6.022\times 10^{23}
(d) 6.022\times 10^{21}

Answers (1)

Answer: A

Solution.

1 mol of sucrose ( C_{12}H_{22}O_{11}) contains = 11\times N_{A} atoms of oxygen,
Here, N_{A} = 6.023\times10^{23}

    1. mol of sucrose ( C_{12}H_{22}O_{11}) contains = 0.01 \times 11 \times N_{A} atoms of oxygen
      = 0.11\times N_{A}=0.11\times N_{A} atoms of oxygen
      =18 g/(1\times2+16)gmol-1
      = 1mol
      1mol of water (H_{2}O)contains 1 \times N_{A} atom of oxygen
      Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water
      = 0.11 N_{A} + 1.0 N_{A}
      ?????=1.11 N_{A}=1.11 N_{A}
      Number of oxygen atoms in solution
      =1.11 \times Avogadro’s number
      =1.11 \times 6.022 \times 10^{23}=6.68\times 10^{23}
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