(solved)Question 1.5 of NCERT Class XI Chemistry Chapter 1

Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.

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Parveen,

Question 1.5
NCERT Class XI Chemistry

Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1

Chemistry Concept for Method I

0.375 Molarity means 1 liter aqueous solution has been obtained by dissolving 0.375 mols of sodium acetate.

However, we need 500 ml. So we have to use 0.375/2 mols of sodium acetate.

Molar mass given as 1 mol = 82.0245 grams.

Hence 0.375/2 mol will be = 0.375/2 x 82.0245 grams.

SF in 0.375 are lower (i.e., 3), so the answer is rounded to 3 digits as 15.4 grams!

Solution by Method I
(but also see Method II after this)

$\displaystyle \text{molarity given } = 0.375 \text{ M}$

$\displaystyle \therefore 1000 \text{ ml contains } 0.375 \text{ mol }$

$\displaystyle \implies 500 \text{ ml contains } \bigg(\frac{0.375}{2}\bigg) \text{ mol }$

$\displaystyle \text{but 1 mol CH}_3\text{COONa} \equiv 82.0245 \text{ g}$

$\displaystyle \therefore \bigg(\frac{0.375}{2}\bigg) \text{ mol } \equiv 82.0245 \times \bigg(\frac{0.375}{2}\bigg) \text{ g} $

$\displaystyle = 15.4 \text{ g (3 Significant Figures)} \:\underline{Ans}$

Video Explanation

Please watch this youtube video for a quick explanation of the solution:

Solution by Method II
(easier and recommended)

$ \begin{aligned} &\text{let moles of CH}_3\text{COONa} = x\\\\ &\frac{\text{moles of solute}}{\text{vol of soln. in liters}} = \text{Molarity}\\\\ &\therefore \frac{x}{(1/2)} = 0.375\text{M (given)}\\\\ &\therefore x = 0.375 \times \frac12\\\\ &= 0.375 \times \frac12 \times 82.0245 \text{ grams}\\\\ &= 15.4 \text{ grams}\:\underline{Ans} \end{aligned} $

Chemistry Concept for Method II

We have used the definition of molarity to obtain the number of moles of sodium acetate. After that we use the given molar mass to convert moles to grams. Note: This method is not only easier to understand, but faster also.

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