Question

How many grams of $40\%$ pure sodium hydroxide is dissolved in $0.5M,250ml$ $NaOH$ solution?
A. $5g$
B. $2g$
C. $12.5g$
D. $4g$

Answer 1

Normal solution we need to know the equivalent $NaOH$, which is determined by partitioning Molecular weight by 1, that is 40 divided by$1=40$. So the equivalent weight of$NaOH$ is 40. To make $1N$ solution, dissolve $40.00g$ of sodium hydroxide in water to make volume 1 liter. For a $0.1N$ solution $4.00g$ of $NaOH$ per liter is required.

Complete step by step solution:
If the solution is in water, then $250mL$ , or $0.25L$ of water will contain $0.25kg$ of water, hence the density of water $\left( {{\rho }_{water}} \right)=1kg{{L}^{-1}}$. The molality of the $NaOH$ is $0.5m$ that is $0.5mol$ of $NaOH$ is dissolved in $1kg$ of water solution. Therefore, $0.25kg$ of water will carry:

$$\Rightarrow \dfrac{x}{0.5}=\dfrac{0.25}{1} \\\ \Rightarrow x=0.125mol$$

So, $0.125$ moles of $NaOH$ are present in the solution. To find out the weight of $NaOH$, we need to find the Molar mass of $NaOH$

$${{M}_{NaOH}}={{M}_{Na}}+{{M}_{O}}+{{M}_{H}} \\\ =23+16+1 \\\ =40\,gmo{{l}^{-1}}$$

Therefore, $0.125$ moles of $NaOH$ will hold: $0.125\times 40=5g$ of $NaOH$
Now, $40\%$ of the solution was used.
Let ‘w’ gram was taken

$$\Rightarrow w\times \dfrac{40}{100}=5 \\\ \Rightarrow w=\dfrac{5\times 10}{4} \\\ \Rightarrow w =12.5 \,g$$

**Hence, the correct option is C.

Additional Information: **
Sodium hydroxide can react violently with strong acids and with water. Other normal names for sodium hydroxide are caustic soda or lye. Sodium hydroxide is the fundamental fixing in family items, for example, fluid channel cleaners. At room temperature, unadulterated sodium hydroxide is a white, scentless strong.

Note:
A molar solution infers concentration regarding moles/liter. One molar (1 M) solution implies one mole of a substance (solute) per liter of solution. A mole implies gram molecular weight or molecular weight of a substance in grams. So the molecular weight of a chemical is likewise its molar weight. To compute the molecular weight one needs to add the atomic weights of the apparent multitude of atoms in the molecular formula unit.