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NCERT Class XII Chemistry
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Question : 30 of 53
Marks: +1, -0
Calculate the amount of benzoic acid (C6H5COOH)(C_6H_5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:  
M=w2×1000/M2×VM= {w_{2}×1000}/{M_{2} × V}
⇒0.15=w2×1000/122×250⇒ 0.15= {w_{2} × 1000}/{122 × 250}
or, w2=1.5×122/4=4.57gw_{2}= {1.5 × 122}/{4}=4.57 {g}
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