Volume of 3 M NaOH (formula weight 40 g ) which can be prepared from 84 g of NaOH is

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Solution

Q.
Volume of 3 M NaOH (formula weight 40 g ${mol}^{- 1}$ ) which can be prepared from 84 g of NaOH is ______ $\times 10^{- 1}$ ${dm}^{3}$ . [2024]

Ans.
(7)

$Moles of NaOH (n_{NaOH}) = \frac{Given mass}{Molar mass} = \frac{84 g}{40 {g mol}^{- 1}} = 2.1 mol$

$Molarity of NaOH \times Volume of NaOH in L = Moles of NaOH$

$3 {mol L}^{- 1} \times Volume of NaOH in L = 2.1 mol$

$Volume of NaOH in L = 0.7 L or 0.7 {dm}^{3} = 7 \times 10^{- 1} {dm}^{3}$

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