JEE MAIN - Chemistry (2023 - 24th January Morning Shift - No. 16)
5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _____________
Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol$$^{-1}$$ respectively
Answer
180
Explanation
Molarity of solution $=\frac{5}{(40)} \frac{(1000)}{(450)}$
$\Rightarrow M \times V=500 \times .1$
$\Rightarrow \frac{5}{40} \times \frac{1000}{450} \times V=500 \times 0.1$
$\mathrm{V}=180 \mathrm{~mL}$
$\Rightarrow M \times V=500 \times .1$
$\Rightarrow \frac{5}{40} \times \frac{1000}{450} \times V=500 \times 0.1$
$\mathrm{V}=180 \mathrm{~mL}$
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