calculate the actual mass of a water molecule in gram.

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Solution

$D e a r S t u d e n t A t o m i c m a s s o f H = 1.007 u A t o m i c m a s s o f O = 15.999 u M o l a r m a s s o f H_{2} O = (2 * 1.007) + 15.999 = 18.013 g / m o l N o w, 6.022 * 10^{23} m o l e c u l e = 1 m o l e \Rightarrow 1 m o l e c u l e = \frac{1}{6.022 * 10^{23}} m o l e S i n c e, 1 m o l e = 18.013 g o f H_{2} O \Rightarrow \frac{1}{6.022 * 10^{23}} m o l e = 18.013 * \frac{1}{6.022 * 10^{23}} g = 2.99 * 10^{- 23} g T h u s, m a s s o f 1 m o l e c u l e o f H_{2} O = 2.99 * 10^{- 23} g R e g a r d s$

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