Question 51
Volume of 3M NaOH (formula weight 40 g mol$$^{-1}$$) which can be prepared from 84 g of NaOH is _____ $$\times 10^{-1}$$ dm$$^3$$.
Correct Answer: 7
Solution
Given: 84 g of NaOH, molar mass = 40 g/mol.
Moles of NaOH = $$\frac{84}{40} = 2.1$$ mol.
For a 3M solution: $$M = \frac{n}{V}$$, so $$V = \frac{n}{M} = \frac{2.1}{3} = 0.7$$ dm$$^3 = 7 \times 10^{-1}$$ dm$$^3$$.
The answer is $$\boxed{7}$$.
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