The number of hydrogen atoms in 0.9 gm glucose, ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ is same as:

Hydrogen atom in given compound,

$\frac{0.9}{180}\times 12$

(a) $\frac{0.048}{32}\times 4$    (b) $\frac{0.17}{17}\times 3$

(c) $\frac{0.30}{30}\times 6$      (d) $\frac{0.03}{2}\times 2$