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Formula of cane sugar is C(12)H(22)O(11)...

Formula of cane sugar is `C_(12)H_(22)O_(11)`. No. of molecules present in 34.2 g of cane sugar is

A

`6.022 xx 10^(21)`

B

`6.022 xx 10^(20)`

C

`6.022 xx 10^(22)`

D

`6.022 xx 10^(18)`

Text Solution

Verified by Experts

The correct Answer is:
C
Gram molecular mass of cane sugar `(C_(12)H_(22)O_(11))`
`= 12xx12 +22 xx 1 +16 xx11`
`= 342.0 g`
342.0 g of can sugar contain
`=6.022 xx 10^(23)` molecules
34.2 g of cane sugar contain
`= (34.2)/(342)xx6.022xx10^(23)`
`=6.022 x 10^(22)` molecules.
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How many surcose molecules (C_(12)H_(22)O_(11)) are present in 3.42g sucrose ? .

Calculate the number of molecules present in 34.2 g of can sugar [C_(12)H_(22)O_(11)] .

Knowledge Check

  • Inversion of cane sugar is a:

    A
    First order reactions
    B
    Zero order reactions
    C
    second order reaction
    D
    Third order reactions

  • The conversion of sugar C_(12)H_(22)O_(11) to CO_(2) is

    A
    Oxidation of Mn in `KMnO_(4)` and production of `Cl_(2)`
    B
    Reduction
    C
    None
    D
    Both

  • Chemically, cane sugar is?

    A
    Lactose
    B
    Fructose
    C
    Sucrose
    D
    Glucose

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    The conversion of sugar C_(12)H_(22)O_(11)toCO_(2) is -

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    Hydrolysis of cane sugar is called

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