3.42g of sucrose is dissolved in 18g $H_{2} O$ in a beaker. The number of oxygen atoms in this solution is:

6.68 \times 10^{23}

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6.09 \times 10^{23}

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6.02 \times 10^{23}

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6.02 \times 10^{21}

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Solution

The correct option is A $6.68 \times 10^{23}$
$n s u r c o s e (C_{12} H_{22} O_{11}) = 3.42 / 342 = 0.01 m o l 1 m o l o f s u c r o s e c o n t a i n s 11 m o l O ∴ 0.01 m o l s u r c o s e w i l l h a v e 11 \times 0.01 m o l = 0.11 m o l O n H_{2} O = 18 / 18 = 1 m o l 1 m o l H_{2} O c o n t a i n s 1 m o l O T o t a l m o l e s o f O = 0.11 + 1.0 = 1.11 m o l N u m b e r o f O a t o m s = 6.02 \times 10^{23} \times 1.11 = 6.68 \times 1 0^{23}$

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Question 9
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) $6.022 \times 10^{21}$

Q. 3.42 g of sucrose are dissolved in 18 g 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}

The number of atoms present in 4.25 g of $({NH}_{3})$ is approximately.

a) $1 \times 10^{23}$

b) $1.5 \times 10^{23}$

c) $0.25 \times 10^{23}$

d) $6.02 \times 10^{23}$

Q. Volume of a mixture of

6.02 \times 10^{23}

oxygen atoms and

3.01 \times 10^{23}

hydrogen molecules at

N T P

is:

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3.42g of sucrose is dissolved in 18g H2O in a beaker. The number of oxygen atoms in this solution is:

The correct option is A 6.68×1023nsurcose(C12H22O11)=3.42/342=0.01mol1molofsucrosecontains11molO∴0.01molsurcosewillhave11×0.01mol=0.11molOnH2O=18/18=1mol1molH2Ocontains1molOTotalmolesofO=0.11+1.0=1.11molNumberofOatoms=6.02×1023×1.11=6.68×1023

3.42g of sucrose is dissolved in 18g $H_{2} O$ in a beaker. The number of oxygen atoms in this solution is: