What weight of sodium hydroxide is required to make 500 cm\(^3\) of 0.2 M solution?
(Na = 23, O = 16, H = 1)
Given:
Volume, V = 500cm\(^3\)
Concentration , C = 0.2 M
n = \(\frac{CV}{1000}\)
n = \(\frac{{0.2}\times{500}}{1000}\)
n = 0.1 mole
Also, recall that n = \(\frac{mass}{Molar mass}\) = \(\frac{m}{M}\)
m = n X M
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
m = 0.1 x 40 = 4g
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