Number of moles of oxygen atoms present in 360 mg of glucose is

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Solution

$D e a r S t u d e n t M a s s o f g l u \cos e = 360 m g = 0.36 g M o l a r m a s s o f g l u \cos e = 180 g / m o l e M o l e s o f g l u \cos e = \frac{0.36 g}{180 g / m o l e} = 0.002 m o l e G l u \cos e - C_{6} H_{12} O_{6} 1 m o l e c u l e o f g l u \cos e c o n t a i n 6 a t o m s o f o x y g e n S o, 1 m o l e o f g l u \cos e c o n t a i n 6 m o l e o f o x y g e n 0.002 m o l e g l u \cos e c o n t a i n o x y g e n = \frac{6 m o l e}{1 m o l e} \times 0.002 m o l e = 0.012 m o l e H e n c e, 360 m g o f g l u \cos e c o n t a i n 0.012 m o l e o f o x y g e n . R e g a r d s$

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